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WORKING PAPER SERIES
Hilbert A-Modules
S. Cerreia– Vioglio, F. Maccheroni, and M. Marinacci
Working Paper n. 544
This Version: March, 2015
IGIER – Università Bocconi, Via Guglielmo Röntgen 1, 20136 Milano –Italy
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Hilbert A-Modules
S. Cerreia-Vioglio, F. Maccheroni, and M. Marinacci
Department of Decision Sciences and IGIER, Università Bocconi
March 2015
Abstract
We consider real pre-Hilbert modules H on Archimedean f -algebras A with
unit e. We provide conditions on A and H such that a Riesz representation
theorem for bounded/continuous A-linear operators holds.
1
Introduction
Let A be an Archimedean f -algebra with (multiplicative) unit e. It is well known that
Archimedean f -algebras are commutative. We next proceed by de…ning the objects we
study in this paper.
De…nition 1 An abelian group (H; +) is an A-module if and only if an outer product
: A H ! H is well de…ned with the following properties, for each a; b 2 A and for
each x; y 2 H:
(1) a (x + y) = a x + a y;
(2) (a + b) x = a x + b x;
(3) a (b x) = (ab) x;
(4) e x = x.
An A-module is a pre-Hilbert A-module if and only if an inner product h ; iH : H
H ! A is well de…ned with the following properties, for each a 2 A and for each
x; y; z 2 H:
We thank Lars Hansen for very useful discussions and comments. Simone Cerreia-Vioglio
and Fabio Maccheroni gratefully acknowledge the …nancial support of MIUR (PRIN grant
20103S5RN3_005). Massimo Marinacci gratefully acknowledges the …nancial support of the AXA
Research Fund.
1
(5) hx; xiH
0, with equality if and only if x = 0;
(6) hx; yiH = hy; xiH ;
(7) hx + y; ziH = hx; ziH + hy; ziH ;
(8) ha x; yiH = a hx; yiH .
For A = R conditions (1)-(4) de…ne vector spaces, while (5)-(8) de…ne pre-Hilbert
spaces. We will use Latin letters a; b; c to denote elements of A, Latin letters x; y; z to
denote elements of H, and Greek letters ; to denote elements of R.
It is well known that1
hx; yi2H
hx; xiH hy; yiH
8x; y 2 H:
We can thus conclude that each z 2 H induces a map f : H ! A, via the formula
f (x) = hx; ziH
8x 2 H;
with the following properties:
- A-linearity f (a x + b y) = af (x) + bf (y) for all a; b 2 A and for all x; y 2 H;
- Boundedness There exists c 2 A+ such that f (x)2
c hx; xiH for all x 2 H.
In light of this fact, we give the following de…nition:
De…nition 2 Let A be an Archimedean f -algebra with unit e and H a pre-Hilbert Amodule. We say that H is self-dual if and only if for each f : H ! A which is A-linear
and bounded there exists y 2 H such that
f (x) = hx; yiH
8x 2 H:
The goal of this paper is to provide conditions on A and H that will allow us to
conclude that a pre-Hilbert A-module H is self-dual. Our initial motivation comes from
Finance. There, Hilbert modules are the extension of the notion of Hilbert spaces that
the analysis of conditional information requires, as …rst shown by Hansen and Richard
[19]. In particular, self-duality is key to represent price operators through traded
stochastic discount factors. Our results provide the general mathematical framework
where conditional asset pricing can be performed.
1
See Huijsmans and de Pagter [23, Theorem 3.4] and also Proposition 4 below.
2
Examples Consider a probability space ( ; F; P ) and assume that G is a sub-algebra of F. Denote by L0 (F) = L0 ( ; F; P ) and L1 (F) = L1 ( ; F; P ), respectively, the space of F-measurable functions and the space of F-measurable and
essentially bounded functions. Similarly, de…ne L0 (G) and L1 (G). De…ne also
L2;0 ( ; G; F; P ) = f 2 L0 (F) : E f 2 jjG 2 L0 (G)
and
L2;1 ( ; G; F; P ) = f 2 L0 (F) : E f 2 jjG 2 L1 (G) :
The inner product, in both cases, can be de…ned by (f; g) 7! E (f gjjG). In Section 6,
we show that L2;0 ( ; G; F; P ) is a pre-Hilbert L0 (G)-module and L2;1 ( ; G; F; P ) is a
pre-Hilbert L1 (G)-module. Both spaces are of particular interest in Finance. The …rst
space is the one originally used in the seminal paper of Hansen and Richard [19]. On the
other hand, in Filipovic, Kupper, and Vogelpoth [14] (see also [30]), the second space
has been shown to represent the family of all continuous and L1 (G)-linear operators
from L2 (F) to L2 (G).2 In other words, L2;1 ( ; G; F; P ) can be interpreted as the
space of all conditional stochastic discount factors (state price densities).
Related literature The literature on self-dual modules can be roughly divided
in two main streams. The …rst one introduced the notion of Hilbert A-modules and
considers complex C -algebras A. In particular, it considers algebras that admit a
concrete representation as a space of continuous functions over a compact space. The
second focuses on a particular algebra of functions, namely, L0 (G) = L0 ( ; G; P ). The
notion of pre-Hilbert A-modules was introduced by Kaplansky [25]. Kaplansky [25]
considers modules over commutative (complex) AW -algebras A with unit and shows
that a pre-Hilbert A-module H is self-dual if and only if H satis…es some extra algebraic
property (De…nition 9). Paschke [33] investigates the properties of self-dual modules
de…ned over complex B -algebras. Two other related papers are Frank [15] and [16]
(see also [29], for a textbook exposition). In both papers, when A is assumed to be a
W complex algebra, a pre-Hilbert A-module H is shown to be self-dual if and only if
the unit ball (properly de…ned) of H is complete with respect to some linear topology.
On the other hand, Guo, in [17] and [18], studies pre-Hilbert L0 (G)-modules H and
shows that they are self-dual if and only if H is complete with respect to a particular
metrizable topology.3
2
L2 (F) = L2 ( ; F; P ) is the space of F-measurable and square integrable functions.
In this paper, we focus on Hilbert modules. For the Banach case, we refer to Cerreia-Vioglio,
Kupper, Maccheroni, Marinacci, and Vogelpoth [11] and the references therein. A pioneer work on
the subject is Haydon, Levy, and Raynaud [20].
3
3
Our Contributions We provide (topological) conditions on A and H that will
allow us to conclude that a pre-Hilbert A-module H is self-dual. We start by considering
A to be an algebra of L1 type (Subsection 2.1). In this case, H can be suitably
topologized with several norm topologies. In particular, two norms stand out: k kH
and k km (Subsection 3.1). When A is of L1 type and H is a pre-Hilbert A-module,
in Theorem 3, we show that the following conditions are equivalent:
(i) H is self-dual;
(ii) BH is “weakly”compact (where BH is the unit ball induced by k kH );
(iii) H is “weakly”sequentially complete;
(iv) BH is complete with respect to k km .
Conditions (ii) and (iii) are novel conditions. On the other hand, a condition of
completeness, similar to Condition (iv), has been found also in the complex case by
Frank [15] (see the proof of [29, Theorem 3.5.1]). When A = R, it is easy to show that
k kH and k km are equivalent (Proposition 9). Thus, in this case, properties (i)-(iv) are
well known to be equivalent and we can conclude that our Theorem 3 is a generalization
of the classical Riesz representation theorem for Hilbert spaces.
We then move to consider A to be an f -algebra of L0 type (Subsection 5.3). In this
case, H can be topologized with an invariant metric dH . When A is of L0 type and
H is a pre-Hilbert A-module, in Theorem 5, we show that the following conditions are
equivalent:
(i’) H is self-dual;
(ii’) H is complete with respect to dH .
We are thus able to obtain Guo’s self-duality result ([17] and [18]). The contribution
to the literature of our Theorem 5 is to show the connection with the self-duality result
for modules on algebras of L1 type. In fact, we show that each pre-Hilbert L0 -module
H contains a dense pre-Hilbert L1 -module He . Thus, verifying the self-duality of H
amounts to verify the self-duality of He , which can then be extended to H via a density
argument.
Outline of the paper Section 2 introduces the two kinds of algebras A we
will consider in studying the self-duality problem. Subsection 2.1 deals with Arens
algebras, that is, real Banach algebras which admit a concrete representation as a
4
space of continuous functions. Algebras of L1 type will belong to this class (De…nition
5). Instead, Subsection 2.2 deals with f -algebras of L0 type (De…nition 6).
In Section 3, we show how a pre-Hilbert A-module naturally turns out to be a vector
space that can also be topologized in several di¤erent and useful ways. In Subsection
3.2, we study the corresponding topological duals.
Section 4 deals with the study of the dual module, that is, the set H of all A-linear
and bounded operators from H to A. The set H turns out to be an A-module which
can also be topologized and its study is key in dealing with the self-duality problem.
From a topological point of view, the structure of H di¤ers depending if A is of L1
type or of L0 type. In Subsection 4.1, we study the …rst case. In Subsection 4.2, we
study the second case. Finally, in Subsection 4.3, we show that H can be identi…ed
with the norm dual of some Banach space when A is of L1 type.
Section 5 contains our results on self-duality. First, we discuss the case when A is
of L1 type. An important subcase is when A is …nite dimensional, which we discuss
right after. We conclude the section by discussing the case in which A is an f -algebra
of L0 type. In Section 6 we discuss …ve examples of pre-Hilbert A-modules that, given
our results, turn out to be self-dual. We relegate to the Appendix the proofs of some
ancillary facts.
2
Function algebras
2.1
Arens algebras
Given a commutative real normed algebra A with multiplicative unit e, we denote by
k kA the norm of A. We denote by A the norm dual of A and by h ; i the dual pairing
of the algebra A, that is, ha; 'i = ' (a) for all a 2 A and ' 2 A . Unless otherwise
speci…ed, the norm dual A of A is endowed with the weak* topology and all of its
subsets are endowed with the relative weak* topology. In the …rst part of the paper,
we will mostly consider commutative real Banach algebras A that admit a concrete
representation. These real Banach algebras were …rst studied by Arens [8] and Kelley
and Vaught [26].4
De…nition 3 A commutative real Banach algebra A with unit e such that
kekA = 1 and kak2A
a2 + b 2
A
8a; b 2 A
is called an Arens algebra.
4
For two more recent studies, see also Albiac and Kanton [2] and [3]. Recall that a Banach algebra
is such that kabkA kakA kbkA for all a; b 2 A.
5
Given an Arens algebra A, de…ne
S = f' 2 A : k'kA = ' (e) = 1g
K = f' 2 S : ' (ab) = ' (a) ' (b)
8a; b 2 Ag :
The set K is compact and Hausdor¤. Denote by C (K) the space of real valued
continuous functions on K. We endow C (K) with the supnorm. It is well known
that A admits a concrete representation, that is, the map T : A ! C (K), de…ned by
T (a) (') = ha; 'i
8' 2 K; 8a 2 A;
is an isometry and an algebra isomorphism (see [26], [2], and [3]).
The cone generated by the squares of A induces a natural order relation on A itself:
a b if and only if a b belongs to the norm closure of fc2 : c 2 Ag. By using standard
techniques, the above concrete representation of A implies that (A; ) is a Riesz space
with strong order unit e and T is also a lattice isomorphism. In particular, K coincides
with the set of all nonzero lattice homomorphisms and A is an Archimedean f -algebra
with unit e. We also have that each element ' 2 K is positive. Finally, k kA is a lattice
norm such that
kakA = min f
0 : jaj
eg and
a2
A
= kak2A
8a 2 A:
In light of these observations, note that for each a
0, there exists a unique b
p
1
2
such that b = a. From now on, we will denote such an element by a 2 or a.
0
Note that if A admits a strictly positive linear functional ' : A ! R, then we could
also renorm A with the norm k k1 : A ! [0; 1), de…ned by kak1 = ' (jaj) for all a 2 A.
It is immediate to see that kak1
k'kA kakA for all a 2 A, and so the k kA norm
topology A is …ner than the k k1 norm topology 1 ; i.e., 1
A . We will denote the
norm dual of A with respect to k k1 by A0 . Finally, we have that A0 A . If A admits
a strictly positive linear functional ' : A ! R, then we could also consider A endowed
with the invariant metric d : A A ! [0; 1), de…ned by d (a; b) = ' (jb aj ^ e) for all
a; b 2 A. It is immediate to see that d (a; b) = ' (jb aj ^ e) ' (jb aj) = kb ak1
for all a; b 2 A, and so the k k1 norm topology 1 is …ner than the d metric topology
d ; i.e., d
1.
The existence of a strictly positive linear functional ' : A ! R will play a key
role in the rest of the paper.5 We conclude the section by exploring the extent of this
assumption and its relation with the existence of a measure m on K whose support
separates the points of A. Before presenting the formal result, we provide a de…nition:
5
Without loss of generality, ' can always be assumed to be such that k'kA = 1.
6
De…nition 4 Let A be an Arens algebra and m a …nite measure on the Borel -algebra
of K. The measure m separates points if and only if the support of m separates the
points of A.
Proposition 1 Let A be an Arens algebra and ' 2 A . The following statements are
equivalent:
(i) The functional ' is strictly positive and such that k'kA = 1;
(ii) There exists a (unique) probability measure m' = m on the Borel -algebra of K
such that suppm = K and
Z
' (a) =
ha; 'i dm (')
8a 2 A;
(1)
K
(iii) There exists a probability measure m' = m that separates points and satis…es (1).
Remark 1 From now on, when we will be dealing with an Arens algebra that admits
a strictly positive linear functional ' on A such that k'kA = 1, the measure m will be
meant to be m' . Viceversa, if A admits a measure that separates points, then ' will
be meant to be de…ned as in (1).
We conclude by de…ning a particular class of Arens algebras which are isomorphic
to some space L1 ( ; G; P ) (see [1, Corollary 2.2]).
De…nition 5 Let A be an Arens algebra. We say that A is of L1 type if and only if A
is Dedekind complete and admits a strictly positive order continuous linear functional
' on A.
2.2
f -algebras
Assume that A is an Archimedean f -algebra with unit e (see Aliprantis and Burkinshaw
[6, De…nition 2.53]). It is well known that e is a weak order unit. If A is Dedekind
1
complete and a
e for some n 2 N, then there exists a unique b 2 A+ such that
n
ab = e. We denote this element a 1 . If a 0 is such that there exists a 1 and b 2 A,
then we alternatively denote ba 1 by b=a. By [22, Theorem 3.9], if A is also Dedekind
complete, for each a
0, there exists a unique b
0 such that b2 = a. Also in this
p
1
case, we will denote such an element by a 2 or a. The principal ideal generated by e
is the set
Ae = fa 2 A : 9 > 0 s.t. jaj
eg :
7
It is immediate to see that Ae is a subalgebra of A with unit e. If A is an Arens algebra,
then Ae = A. If there exists a linear and strictly positive functional ' : Ae ! R, then
we can de…ne d : A A ! [0; 1) by
d (a; b) = ' (jb
aj ^ e)
8a; b 2 A:
As in the case of an Arens algebra, d is an invariant metric. As already noted, an
Arens algebra, in particular one of L1 type, is an Archimedean f -algebra with unit. In
this paper, other than algebras of L1 type, we focus also on another particular class
of f -algebras:
De…nition 6 Let A be an Archimedean f -algebra with unit e. We say that A is an f algebra of L0 type if and only if Ae is an Arens algebra of L1 type and A is Dedekind
complete and d complete.
By [6, Theorems 2.28 and 4.7], if A is an f -algebra of L0 type, d is generated by the
Riesz pseudonorm c 7! ' (jcj ^ e), then it is easy to prove that the topology generated
by d is linear, locally solid, and Fatou. Moreover, it can be shown that A is universally
complete and such that:
d
1. If an # 0 and b
2. If b
3
0, then an b # 0 and an b ! 0;
d
d
0 and an ! a, then ban ! ba.
The vector space structure of H
In this section, we will …rst show that a pre-Hilbert A-module has a natural structure
of vector space. Next, we will show that the A valued inner product h ; iH shares most
of the properties of standard real valued inner products. In particular, under mild
assumptions on A, we will show that it also induces a real valued inner product on H,
thus making H into a pre-Hilbert space.
We use the outer product to de…ne a scalar product:
e
We next show that
e
: R H ! H
:
( ; x) 7! ( e) x
makes the abelian group H into a real vector space.
Proposition 2 Let A be an Archimedean f -algebra with unit e and H an A-module.
(H; +; e ) is a real vector space.
Proof. By assumption, H is an abelian group. For each ;
we have that
8
2 R and each x; y 2 H,
(1)
e
e
(x + y) = e (x + y) = ( e) x + ( e) y =
e
(2) ( + )
e
x+
e
y;
x = (( + ) e) x = ( e + e) x = ( e) x + ( e) x =
(3)
e
(
(4) 1
e
x = (1e) x = e x = x.
x) = ( e) (( e) x) = (( e) ( e)) x = ((
From now on, we will often write x in place of
e
) e) x = (
)
e
x+
e
x;
e
x;
x.
Corollary 1 Let A be an Archimedean f -algebra with unit e and H an A-module. If
f : H ! A is an A-linear operator, then f is linear.
If A is an Arens algebra, given a …nite probability measure m on the Borel -algebra
of K we can also de…ne h ; im : H H ! R by
Z
hx; yim =
hhx; yiH ; 'i dm (') 8x; y 2 H:
K
For each ' 2 K, we also de…ne and study the functionals h ; i' : H
by
hx; yi' = hhx; yiH ; 'i
8x; y 2 H:
H ! R de…ned
Note that h ; i' = h ; i ' for all ' 2 K where ' is the Dirac measure at '. We next
show that h ; im is a symmetric bilinear form which is positive semide…nite on H H.
Proposition 3 Let A be an Arens algebra and H a pre-Hilbert A-module. The following statements are true:
1. h ; im is a positive semide…nite symmetric bilinear form;
2. hx; xim = 0 implies x = 0, provided m separates points;
3. hx; yi2m
hx; xim hy; yim for all x; y 2 H;
4. hx; a yim = ha x; yim for all a 2 A and for all x; y 2 H;
5. hx; yim = ' (hx; yiH ) for all x; y 2 H, provided m separates points.
Proof. We here prove points 2. and 5. and leave the remaining easy ones to the
reader. Assume that m separates points. De…ne ' as in (1). By Proposition 1, it
follows that
Z
' (a) =
ha; 'i dm (')
8a 2 A
(2)
K
is a strictly positive and linear functional.
9
5. By de…nition of h ; im and (2), observe that for each x; y 2 H
Z
hx; yim =
hhx; yiH ; 'i dm (') = ' (hx; yiH ) :
K
2. By assumption, hx; xiH
0 for all x 2 H. By point 5., if hx; xim = 0, then
' (hx; xiH ) = 0. Since ' is strictly positive, we have that hx; xiH = 0, proving that
x = 0.
Corollary 2 Let A be an Arens algebra and H a pre-Hilbert A-module. If A admits a
measure m that separates points, then (H; +; e ; h ; im ) is a pre-Hilbert space.
Proposition 4 Let A be an Arens algebra and H a pre-Hilbert A-module. The following statements are true:
1. hx; yi2H
hx; xiH hy; yiH for all x; y 2 H;
1
2. jhx; yiH j
3.
hx; yi2H
1
hx; xiH2 hy; yiH2 for all x; y 2 H;
A
khx; xiH kA khy; yiH kA for all x; y 2 H;
4. khx; yiH kA
khx; xiH kA2 khy; yiH kA2 for all x; y 2 H.
1
1
Proof. By Proposition 2 and Corollary 1 and since A is, in particular, an Archimedean
f -algebra with unit, point 1. is an easy consequence of [23, Theorem 3.4]. Since A is
an Arens algebra, each positive element admits a unique square root and point 2. also
follows. Since A is an Arens algebra and k kA is also a lattice norm, we have that for
each x; y 2 H
khx; yiH k2A = hx; yi2H
A
khx; xiH hy; yiH kA
khx; xiH kA khy; yiH kA ;
proving points 3. and 4.
Remark 2 If A is a Dedekind complete Archimedean f -algebra with unit e, then points
1. and 2. are still true and their proofs remain the same.
3.1
Topological structure
Since a pre-Hilbert A-module H is also a vector space, we can try to endow H with a
topology induced by either a norm or an invariant metric. In fact, given the structure
of A and H, we have several di¤erent competing norms and topologies. The next
subsections are devoted to the study of these norms and metric and their relations.
Before starting, note that if A is an Arens algebra or a Dedekind complete Archimedean
10
f -algebra with unit, then h ; iH de…nes a vector-valued norm,6 N : H ! A+ , via the
formula
1
N (x) = hx; xiH2
8x 2 H:
If A were equal to R, then N would be a standard norm and we would say that
def
xn ! x () N (x
xn ) ! 0:
Since R is always endowed with the usual topology, this de…nition would be unambiguous. When A 6= R, such a statement is not true anymore, since we could endow A with
di¤erent linear topologies changing the meaning of N (x xn ) ! 0. In other words,
by combining the topological structure of A with N we are able to induce di¤erent
topologies on H (see point 5. of Proposition 5, point 3. of Proposition 6, equation (5),
and point 2. of Proposition 7).
3.1.1
The k kH norm
Assume A is an Arens algebra. De…ne k kH : H ! [0; 1) by
q
8x 2 H:
kxkH = khx; xiH kA
For each ' 2 K, de…ne also k k' : H ! [0; 1) by
kxk' =
q
hx; xi'
8x 2 H:
Proposition 5 Let A be an Arens algebra and H a pre-Hilbert A-module. The following statements are true:
1. k k' is a seminorm for all ' 2 K;
2. k kH is a norm;
3. kxkH = max'2K kxk' =
4. ka xkH
q
max'2K hx; xi' for all x 2 H;
kakA kxkH for all a 2 A and all x 2 H;
5. kxkH = kN (x)kA for all x 2 H.
6
In particular, N is such that
1. N (x) = 0 if and only if x = 0;
2. N (a x) = jaj N (x) for all a 2 A and for all x 2 H;
3. N (x + y)
N (x) + N (y) for all x; y 2 H.
11
Proof. Points 1. and 2. follow from routine arguments.
3. Recall that T : A ! C (K) is a linear isometry. Thus, we have that
kxk2H = khx; xiH kA = kT (hx; xiH )k1 = max jT (hx; xiH ) (')j = max jhhx; xiH ; 'ij
'2K
= max hx; xi' = max hx; xi' =
'2K
'2K
'2K
max kxk2'
'2K
8x 2 H;
proving the statement.
4. Given any a 2 A and x 2 H, it holds
ka xk2H = kha x; a xiH kA = a2 hx; xiH
a2
A
A
khx; xiH kA
kak2A khx; xiH kA
proving the statement.
p
p
5. Since A is an Arens algebra, it follows that k akA = kakA for all a 2 A+ .7
This implies that
q
q
kxkH = khx; xiH kA =
hx; xiH = kN (x)kA
8x 2 H;
A
proving the statement.
By Proposition 4, it readily follows that
khx; yiH kA
kxkH kykH
8x; y 2 H:
(3)
Corollary 3 Let A be an Arens algebra and H a pre-Hilbert A-module. For each
y 2 H, the functional h ; yiH : H ! A is A-linear, k kH k kA continuous, and has
norm kykH .
Proof. Fix y 2 H. It is immediate to see that the operator induced by y is A-linear,
thus, linear. Continuity easily follows from (3). Since the norm of the linear operator
is given by
sup khx; yiH kA = kxkH : x 6= 0 ;
the statement easily follows from (3) and the de…nition of k kH .
In light of these observations and since k kH can be de…ned for any pre-Hilbert
A-module when A is an Arens algebra, we propose the following de…nition:
De…nition 7 Let A be an Arens algebra and H a pre-Hilbert A-module. We say that
H is an Hilbert A-module if and only if H is k kH complete.
7
1
Recall that if a 0, then a 2 = b is the unique positive element such that b2 = a. Since A is an
p 2
2
Arens algebra, it follows that kakA = b2 A = kbkA = k akA , proving that
q
p
a A = kakA
8a 2 A+ :
12
3.1.2
The k kp norm
Assume A is an Arens algebra that admits a strictly positive linear functional ' such
that k'kA = 1. De…ne k kp : H ! [0; 1) by
1
kxkp = ' hx; xiH2
8x 2 H:
Proposition 6 Let A be an Arens algebra and H a pre-Hilbert A-module. If A admits
a strictly positive functional ' such that k'kA = 1, then the following statements are
true:
1. k kp is a norm;
2. ka xkp
kakA kxkp for all a 2 A and for all x 2 H;
3. kxkp = ' (N (x)) = kN (x)k1 for all x 2 H.
Proof. Since ' is strictly positive, note that
1
1
x = 0 () hx; xiH = 0 () hx; xiH2 = 0 () ' hx; xiH2 = 0 () kxkp = 0:
Second, since ' is linear, observe that for each
2 R and for each x 2 H
1
1
1
k xkp = ' h x; xiH2 = ' j j hx; xiH2 = j j ' hx; xiH2 = j j kxkp :
Third, by Proposition 4, note that for each x; y 2 H
1
hx + y; x + yiH = hx; xiH + 2 hx; yiH + hy; yiH
1
2
1
= hx; xiH2 + hy; yiH2
1
hx; xiH + 2 hx; xiH2 hy; yiH2 + hy; yiH
:
1
1
1
We can conclude that hx + y; x + yiH2
hx; xiH2 + hy; yiH2 for all x; y 2 H. Since ' is
positive and linear, the triangular inequality for k kp follows.
2. Given any a 2 A and x 2 H, since ' is positive and linear, it holds
1
1
ka xkp = ' ha x; a xiH2 = ' jaj hx; xiH2
1
1
' (kakA e) hx; xiH2
kakA ' hx; xiH2 = kakA kxkp ;
proving the statement.
3. The statement follows by de…nition of k kp and k k1 .
13
The k km norm
3.1.3
Assume A is an Arens algebra that admits a strictly positive linear functional ' such
that k'kA = 1 or, equivalently, a probability measure m that separates points (see
Proposition 1). De…ne k km : H ! [0; 1) by
sZ
q
kxkm = hx; xim =
hx; xi' dm (')
8x 2 H:
(4)
K
By Propositions 1 and 3, h ; im is an inner product on H and it is immediate to see
that k km is a norm and
q
q
8x 2 H:
(5)
kxkm = ' (hx; xiH ) = ' N (x)2
The dH metric
3.1.4
Assume that A is either a Dedekind complete Archimedean f -algebra with unit that
admits a strictly positive linear functional ' : Ae ! R or A is an Arens algebra that
admits a strictly positive linear functional ' : A ! R. In the second case, assume also
that k'kA = 1. De…ne dH : H H ! [0; 1) by
dH (x; y) = ' (N (x
y) ^ e)
8x; y 2 H:
(6)
Recall that the hypothesis of Dedekind completeness is needed to de…ne N (x) =
1
hx; xiH2 when A is an Archimedean f -algebra with unit.
Proposition 7 Let A be either a Dedekind complete Archimedean f -algebra with unit
e or an Arens algebra and H a pre-Hilbert A-module. If A admits a strictly positive
functional ' : Ae ! R, then the following statements are true:
1. dH is an invariant metric;
2. dH (x; y) = d (0; N (x
y)) for all x; y 2 H.
Proof. 1. Since ' is strictly positive, we have that
dH (x; y) = 0 () ' (N (x
() N (x
y) ^ e) = 0 () N (x
y) ^ e = 0
y) = 0 () x = y:
It is immediate to see that dH (x; y) = dH (y; x) for all x; y 2 H as well as
dH (x + z; y + z) = dH (x; y)
14
8x; y; z 2 H:
Finally, by [6, Lemma 1.4] and since N (x + y)
can conclude that
dH (x; y) = ' (N (x
' ((N (x
y) ^ e) = ' (N ((x
z) + (z
y)) ^ e)
z) + N (z
N (x) + N (y) for all x; y 2 H, we
y)) ^ e)
' (N (x
z) ^ e + N (z
y) ^ e)
8x; y; z 2 H;
= dH (x; z) + dH (z; y)
proving the statement.
2. By de…nition of dH , d, and N , we have that
y) ^ e) = ' (jN (x
dH (x; y) = ' (N (x
y)
0j ^ e) = d (0; N (x
y))
8x; y 2 H;
proving the statement.
3.1.5
Relations among norms
Assume A is an Arens Algebra that admits a strictly positive linear functional ' such
that k'kA = 1. First, by Propositions 1 and 5 and by equation (4), we have that
kxkm
kxkH
8x 2 H:
We can conclude that
kk
kk
xn !H 0 =) xn !m 0:
The k kH norm topology H is thus …ner than the k km norm topology m ; i.e., m
Similarly, by Proposition 1, we have that
Z D
Z
E
1
1
1
2
2
kxkp = ' hx; xiH =
hx; xiH ; ' dm (') =
hhx; xiH ; 'i 2 dm (')
K
K
sZ
K
hhx; xiH ; 'i dm (') = kxkm
We can conclude that
8x 2 H:
k kp
kk
xn !m 0 =) xn ! 0:
The k km norm topology m is thus …ner than the k kp norm topology
Summing up, we have that
kxkp
kxkm
H.
(7)
p;
i.e.,
p
m.
kxkH
8x 2 H:
(8)
ykp
8x; y 2 H:
(9)
Finally, note that
dH (x; y)
kx
The k kp norm topology p is thus …ner than the dH topology dH ; i.e., dH
p.
We next explore the continuitynboundedness properties of A-linear operators: f :
H ! A. We then conclude by showing that our three norms are equivalent when A is
…nite dimensional.
15
Proposition 8 Let A be a Dedekind complete Arens algebra and H a pre-Hilbert Amodule. If A admits a strictly positive linear functional ' such that k'kA = 1, the
following statements are true:
1. An A-linear operator f : H ! A is bounded if and only if f is k kH
continuous;
k kA
2. If f : H ! A is A-linear and k kH
k kp k k1 continuous;
k kA continuous, then f : H ! A is
3. If f : H ! A is A-linear and k kH
k km k k1 continuous;
k kA continuous, then f : H ! A is
4. For each x; y 2 H,
j' (hx; yiH )j
' (jhx; yiH j) = khx; yiH k1
kxkp kykH ;
' (jhx; yiH j) = khx; yiH k1
kxkm kykH :
5. For each x; y 2 H,
j' (hx; yiH )j
Proof. 1. By Corollary 1 and since f is A-linear, f is linear. If f is bounded, then
there exists c 2 A+ such that
f 2 (x)
c hx; xiH
8x 2 H:
Since k kA is a lattice norm and A is an Arens algebra, this implies that
kf (x)k2A = f 2 (x)
A
kc hx; xiH kA
kckA khx; xiH kA
8x 2 H;
p
that is, kf (x)kA
kckA kxkH for all x 2 H. We can conclude that f is k kH k kA
continuous. Viceversa, assume that f is k kH k kA continuous. It follows that there
exists k 0 such that kf (x)kA k kxkH for all x 2 H. Fix x 2 H. Since f is A-linear,
it follows that for each a 2 A
a2 f 2 (x)
A
= kaf (x)k2A = kf (a x)k2A
k 2 kha x; a xiH kA = k 2 a2 hx; xiH
Since A is a Dedekind complete Arens algebra, this implies that f 2 (x)
Since x was arbitrarily chosen, it follows that f is bounded.
8
A
:
(k 2 e) hx; xiH .8
For a Dedekind complete Arens algebra A, it is true that if c; d 2 A+ are such that
kbckA
kbdkA
8b 2 A+ ;
then c d. In our case, c = f 2 (x), d = k 2 hx; xiH , and, given Subsection 2.1, it is enough to observe
that A+ = a2 : a 2 A .
16
2. By point 1. and since A is a Dedekind complete Arens algebra, if f : H ! A is
k kH k kA continuous, then there exists c 2 A+ such that
1
jf (x)j
c hx; xiH2
8x 2 H:
Since ' is (strictly) positive, we have that
1
1
0
kf (x)k1 = ' (jf (x)j)
1
' c hx; xiH2
= kckA ' hx; xiH2 = kckA kxkp
' (kckA e) hx; xiH2
8x 2 H:
3. By the proof of point 2. and (7), the statement follows.
4. By Proposition 4 and since ' is (strictly) positive and linear, we have that
1
j' (hx; yiH )j
1
1
1
' (jhx; yiH j) ' hx; xiH2 hy; yiH2
' hx; xiH2
hy; yiH2
q
1
= khy; yiH kA ' hx; xiH2 = kykH kxkp
8x; y 2 H;
e
A
proving the statement, since ' (jhx; yiH j) = khx; yiH k1 for all x; y 2 H.
5. By the proof of point 4. and (8), the statement follows.
Remark 3 It is important to note that in the proof of point 1. the existence of a
strictly positive functional ' did not play any role. Similarly, in the proof of points 4.
and 5. the assumption of Dedekind completeness was not used.
Proposition 9 Let A be an Arens algebra and H a pre-Hilbert A-module. If A is …nite
dimensional, then A admits a strictly positive linear functional ' such that k'kA = 1
and the norms k kp , k km , and k kH are equivalent.
Proof. Since A is …nite dimensional, we have that K is …nite (see Semadeni [36,
Corollary 6.4.9 and Propositions 6.2.10 and 7.1.4]). Since K is …nite, consider m = m'
1
such that m (f'g) = jKj
for all ' 2 K. By Proposition 1, ' is strictly positive, linear,
and such that k'kA = 1. Consider now a generic strictly positive linear functional '.
By Proposition 1 and since K is …nite, it is immediate to see that m (f'g) > 0 for all
' 2 K. Consider a sequence fxn gn2N H. By Proposition 1, Proposition 6, point 3.
of Proposition 5, and since K is …nite, it follows that
Z D
E
1
k kp
xn ! 0 () kxn kp ! 0 ()
hxn ; xn iH2 ; ' dm (') ! 0
K
kk
() kxn k' ! 0 8' 2 K () kxn kH ! 0 () xn !H 0;
proving that k kp and k kH are equivalent. By (8), the statement follows.
17
3.2
The norm duals of H
Assume A is an Arens algebra that admits a strictly positive functional ' such that
k'kA = 1. Recall that on H we have at least two competing norms for H: k kH and
k km . We denote by BH the closed unit ball with respect to k kH and by Bm the closed
unit ball with respect to k km .
In this subsection, we study the norm duals these norms induce. We denote them,
respectively, H and H 0 . Since kxkm
kxkH for all x 2 H, we have that H 0
H
and, given l 2 H 0 , that
klkH 0 = sup fjl (x)j : kxkm
sup fjl (x)j : kxkH
1g
1g = klkH :
(10)
Consider ' 2 A and y 2 H. By Corollary 1, note that ' h ; yiH is a linear
functional. By Corollary 3, the k kH continuity of ' h ; yiH follows, since the operator
h ; yiH is k kH k kA continuous and ' is k kA continuous. On the other hand, by
Proposition 8 (see also Remark 3), if ' 2 A0 A , k km continuity of ' h ; yiH follows
since the operator h ; yiH is k km k k1 continuous and ' is k k1 continuous.
Lemma 1 Let A be an Arens algebra and H a pre-Hilbert A-module. If A admits a
strictly positive functional ' such that k'kA = 1 and S : H ! H 0 H is de…ned by
8y 2 H;
S (y) = ' (h ; yiH )
then:
1. S is well de…ned and linear;
2. ker S = f0g;
3. S is such that kS (y)kH 0 = kykm for all y 2 H. In particular, S is k km
continuous.
k kH 0
Proof. 1. Note that ' 2 A0
A . By the arguments preceding the proof, we have
that for each y 2 H the functional S (y) is linear and k kH and k km continuous, so
that, S is well de…ned. For each 1 ; 2 2 R and y1 ; y2 2 H, we also have that
S(
1 y1
+
2 y2 ) (x)
= ' (hx;
= ' (hx; (
2 y2 iH )
+
1 e)
1
hx; y1 iH +
=
1 ' (hx; y1 iH )
=
1S
2
+
(y1 ) (x) +
18
+(
1 y1 iH
= ' (hx;
y1 iH + hx; (
1 e) hx; y1 iH
= ' ((
= '(
1 y1
2 e)
+ hx;
y2 iH )
2 e) hx; y2 iH )
hx; y2 iH )
2 ' (hx; y2 iH )
2S
(y2 ) (x)
8x 2 H;
2 y2 iH )
proving S is linear.
2. Consider y 2 H. Assume that S (y) = 0. It follows that
8x 2 H:
' (hx; yiH ) = S (y) (x) = 0
By choosing x = y and since ' is strictly positive, we have that
' (hy; yiH ) = 0 =) hy; yiH = 0 =) y = 0:
This yields that ker S = f0g.
3. By Corollary 2 (see also Proposition 3), recall that H with the inner product
h ; im is a pre-Hilbert space. It follows that
jS (y) (x)j = j' (hx; yiH )j = jhx; yim j
kxkm kykm
8x; y 2 H:
We can conclude that kS (y)kH 0 = kykm for all y 2 H.
Next proposition shows that S : H ! H is an isometry when H is endowed with
the norm k kp and H is endowed with the norm k kH .
Proposition 10 Let A be a Dedekind complete Arens algebra and H a pre-Hilbert Amodule. If A admits a strictly positive linear functional ' such that k'kA = 1, then
kS (y)kH = kykp for all y 2 H.
Proof. Consider l 2 S (H). It follows that there exists y 2 H such that l (x) =
' (hx; yiH ) for all x 2 H and
klkH = sup ' (hx; yiH ) :
x2BH
First, consider the problem supx2BH hx; yiH . If x 2 BH , then khx; xiH kA
1
2
yields hx; xiH
1 which
e. By Proposition 4, this implies that
1
hx; yiH
1
1
hx; xiH2 hy; yiH2
hy; yiH2
8x 2 BH :
1
Since A is Dedekind complete, it follows that A 3 supx2BH hx; yiH
1
we show that kykp = ' hy; yiH2
1
2
hy; yiH2 . Next,
= supx2BH ' (hx; yiH ) = klkH . Consider an =
1
hy; yiH + n1 e 0 for all n 2 N. Since hy; yiH2 2 A+ , it is immediate to see that an is
invertible for all n 2 N. De…ne yn = an 1 y for all n 2 N.
Claim: For each n 2 N,
hyn ; yn iH = an 1 y; an 1 y
19
H
e:
1
Moreover, 0
1
kk
hy; yiH2 for all n 2 N, hyn ; yiH ", and hyn ; yiH !A hy; yiH2 .
hyn ; yiH
Proof of the Claim. See the Appendix.
We can conclude that kyn kH
1
2
hy; yiH
supx2BH hx; yiH
1 for all n 2 N. Since ' 2 A is positive and
hz; yiH for all z 2 BH , it follows that
1
' hy; yiH2
'
sup hx; yiH
x2BH
' (hz; yiH )
8z 2 BH ;
1
that is, ' hy; yiH2
supx2BH ' (hx; yiH ). Viceversa, since ' 2 A is positive, note
1
supn2N ' (hyn ; yiH ) = limn ' (hyn ; yiH ) = ' hy; yiH2 , prov-
that supx2BH ' (hx; yiH )
ing the statement.
Remark 4 Observe that, under the assumptions of previous proposition, the completion of H with respect to k kp , denoted by Hp , can be identi…ed with the k kH closure
of S (H) in H . We will always adopt this identi…cation.
We conclude the section with an ancillary lemma which will be instrumental in
proving one of our main results on self-duality.
Lemma 2 Let A be a Dedekind complete Arens algebra and H a pre-Hilbert A-module.
If A admits a strictly positive functional ' such that k'kA = 1, then h ; im = ' h ; iH
admits a unique bilinear extension from H Hp to R, denoted h ; im , such that:
1. hx; yim
kxkH kykp for all x 2 H and y 2 Hp ;
2. If hx; yim = hx; y 0 im for all x 2 H, then y = y 0 .
Proof. Denote the dual pairing of H and H by h ; iH;H . Note that for each x; y 2 H
hx; S (y)iH;H = S (y) (x) = hx; yim :
Moreover, by Proposition 10, hx; S (y)iH;H
kxkH kS (y)kH = kxkH kykp for all
x; y 2 H. Since Hp can be identi…ed with clk kH (S (H)) ; k kH , by de…ning h ; im =
h ; iH;H , the main statement and point 1. follow. We next prove uniqueness. Assume
that h ; im is a bilinear extension of h ; im satisfying 1. Consider y 2 Hp and x 2 H.
There exists fyn gn2N
hx; yim
k kp
H such that yn ! y. Thus,
hx; yim
hx; yim
hx; yn im + jhx; yn im
hx; yim j
= hx; yim
hx; yn im + jhx; yn im
hx; yim j ! 0;
proving that hx; yim = hx; yim . Since x and y were arbitrarily chosen, uniqueness
follows. Finally, we have that if y; y 0 2 Hp are such that for each x 2 H
hx; yim = hx; y 0 im ;
then, hx; yiH;H = hx; y 0 iH;H , yielding y = y 0 .
20
4
Dual module
Given an Archimedean f -algebra A with unit e and a pre-Hilbert A-module H, we
de…ne
H = f 2 AH : f is A-linear and bounded :
Proposition 11 If A is a Dedekind complete Archimedean f -algebra with unit e and
H a pre-Hilbert A-module, then H is an A-module.
Proof. De…ne + : H
H ! H to be such that for each f; g 2 H
(f + g) (x) = f (x) + g (x)
8x 2 H:
In other words, + is the usual pointwise sum of operators. De…ne : A
to be such that for each a 2 A and for each f 2 H
(a f ) (x) = af (x)
H
!H
8x 2 H:
It is immediate to verify that H is closed under + and .9 In particular, (H; +) is an
abelian group. Note that for each a; b 2 A and each f; g 2 H :
1. (a (f + g)) (x) = a ((f + g) (x)) = a (f (x) + g (x)) = af (x)+ag (x) = (a f ) (x)+
(a g) (x) = (a f + a g) (x) for all x 2 H, that is, a (f + g) = a f + a g.
2. ((a + b) f ) (x) = (a + b) f (x) = af (x) + bf (x) = (a f ) (x) + (b f ) (x) =
(a f + b f ) (x) for all x 2 H, that is, (a + b) f = a f + b f .
3. (a (b f )) (x) = a ((b f ) (x)) = a (bf (x)) = (ab) f (x) = ((ab) f ) (x) for all
x 2 H, that is, a (b f ) = (ab) f .
4. (e f ) (x) = ef (x) = f (x) for all x 2 H, that is, e f = f .
If A is Dedekind complete, then de…ne S : H ! H by
S (y) = h ; yiH
8y 2 H:
Given Remark 2 and the properties of h ; iH , the map S is well de…ned. We next
study the (topological) properties of this map and its connection to the self-duality
problem. Since the topologies involved are di¤erent, we split this study in two cases:
Dedekind complete Arens algebras and f -algebras of L0 type. Before starting, we need
one more de…nition:
9
Dedekind completeness yields the existence of the square root for positive elements of A. This
property is used to show that the sum of two bounded A-linear operators is bounded.
21
De…nition 8 Let H1 and H2 be two A-modules and S : H1 ! H2 . We say that S
is a module homomorphism if and only if
S (a x + b y) = a S (x) + b S (y)
We say that S
morphism.
4.1
8a; b 2 A; 8x; y 2 H1 :
is a module isomorphism if and only if it is a bijective module homo-
Dual module: Arens algebras
If A is a Dedekind complete Arens algebra (see Proposition 8 and Remark 3), we have
that
H = f 2 AH : f is A-linear and k kH k kA continuous :
In this case, we de…ne k kH : H ! [0; 1) by
kf kH = sup kf (x)kA
x2BH
8f 2 H :
Recall that if f 2 H , then f is linear. Thus, in this case, we have that H
B (H; A),
where the latter is the set of all norm bounded linear operators from H to A when H
is endowed with k kH and A is endowed with k kA .
Proposition 12 Let A be a Dedekind complete Arens algebra and H a pre-Hilbert
A-module. The following statements are true:
1. H = f 2 AH : f is A-linear and k kH
2. H
k kA continuous .
is a k kH complete A-module.
3. S is a well de…ned module homomorphism and kS (y)kH
y 2 H.
4. If H is self-dual, then S
= kykH for all
is onto and H is k kH complete.
Proof. 1. It follows from point 1. of Proposition 8 (see also Remark 3).
2. By Proposition 11, H is an A-module. In particular, H is a vector subspace
of B (H; A). Consider a k kH Cauchy sequence ffn gn2N
H
B (H; A). By [4,
kk
H
Theorem 6.6], we have that there exists f 2 B (H; A) such that fn !
f . We are left
to show that f is A-linear. First, observe that f : H ! A is such that
f (x) = lim fn (x)
n
22
8x 2 H
where the limit is in k kA norm. We can conclude that for each a; b 2 A and x; y 2 H
kk
kk
kk
kk
fn (x) !A f (x) ; fn (y) !A f (y) =) afn (x) !A af (x) ; bfn (y) !A bf (y)
kk
=) afn (x) + bfn (y) !A af (x) + bf (y) :
kk
At the same time, afn (x)+bfn (y) = fn (a x + b y) !A f (a x + b y) for all a; b 2 A
and x; y 2 H. By the uniqueness of the limit, we can conclude that f (a x + b y) =
af (x) + bf (y) for all a; b 2 A and x; y 2 H, proving the statement.
3. De…ne S : H ! H by
S (y) (x) = hx; yiH
8x 2 H:
By Corollary 3, it follows that S is well de…ned and such that kS (y)kH = kykH
for all y 2 H. Note also that for each a; b 2 A and for each y; z 2 H
S (a y + b z) (x) = hx; a y + b ziH = a hx; yiH + b hx; ziH
= aS (y) (x) + bS (z) (x)
8x 2 H;
= (a S (y)) (x) + (b S (z)) (x)
in other words, we have that S (a y + b z) = a S (y) + b S (z), that is, S is a
module homomorphism. We can also conclude that
kS (y)
S (z)kH = kS (y
z)kH = ky
zkH ;
that is, S is an isometry.
4. If H is self-dual, it is immediate to see that S is onto. Consider a k kH Cauchy
sequence fxn gn2N
H. Since S is an isometry, it follows that fS (xn )gn2N is a
k kH Cauchy sequence in H . Since H is k kH complete and S is onto, it follows
that there exists f 2 H
kk
H
such that S (xn ) !
f = S (x) for some x 2 H . Since
kk
S is an isometry, we have that xn !H x, proving that H is k kH complete.
4.2
Dual module: f -algebras of L0 type
In order to discuss the continuity properties of the map S we need to endow H with
a topology. We saw that if A is an Arens algebra, then the choice of topology for H
is rather natural: the one induced by the standard operator norm k kH . The same
choice cannot be made when A is an f -algebra of L0 type. Thus, …rst we de…ne the
operator vector-valued norm N : H ! A+ by
N (f ) = sup sup
x2H
n2N
jf (x)j
N (x) + n1 e
23
8f 2 H :
Then, we de…ne the metric dH : H
H ! [0; 1)
dH (f; g) = d (0; N (f
8f; g 2 H :
g))
Lemma 3 If A is an f -algebra of L0 type and H a pre-Hilbert A-module, then dH is
an invariant metric.
Proposition 13 Let A be an f -algebra of L0 type and H a pre-Hilbert A-module. The
following statements are true:
1. H = f 2 AH : f is A-linear and dH
d continuous .
2. S is a well de…ned module homomorphism and dH (S (x) ; S (y)) = dH (x; y)
for all x; y 2 H.
3. If H is self-dual, then S
is onto and H is dH complete.
Proof. 1. We …rst show that
f 2 AH : f is A-linear and dH
H
d continuous :
Consider f 2 H . We only need to show that f is dH
bounded, there exists c 2 A+ such that
jf (x)j
cN (x)
d continuous. Since f is
8x 2 H:
(11)
By [35, Theorem 1.32] and since f is A-linear, f is linear and we only need to show
dH
continuity at 0. Consider fxn gn2N
H such that xn !
0. By de…nition of dH , it
d
d
follows that N (xn ) ! 0, thus cN (xn ) ! 0. By (11) and de…nition of d, it follows
that d (0; jf (xn )j) = ' (jf (xn )j ^ e) ' ((cN (xn )) ^ e) = d (0; cN (xn )) ! 0, proving
continuity at 0.
As for the opposite inclusion, we consider an A-linear and dH d continuous function
f and show it is bounded. First, de…ne BHe = fx 2 H : N (x) eg. Consider a
sequence fxn gn2N
BHe and f n gn2N
R such that n ! 0. It follows that there
exists n 2 N such that j n j < 1 for all n n. This implies that for n n
dH (0;
n xn )
= ' (N (
' (j
n xn )
nj e
^ e) = ' ((j
^ e) = j
n j ' (e)
nj N
(xn )) ^ e)
! 0;
d
H
proving that n xn !
0. By [35, Theorem 1.30] and [35, Theorem 1.32], we have that
BHe is topologically bounded and so are f (BHe ) and fjf (x)jgx2BHe . Consider the
following binary relation on BHe :
x
y () jf (x)j
24
jf (y)j :
It is immediate to see that
is re‡exive and transitive. Next, consider x; y 2 BHe .
0
Since A is an f -algebra of L type, it follows that there exists c1 ; c2 2 A+ such that
c1 ^ c2 = 0, c1 ; c2 e, and
c1 jf (x)j + c2 jf (y)j = jf (x)j _ jf (y)j :10
De…ne z = c1 x+c2 y. Note that N (c1 x + c2 y) c1 N (x)+c2 N (y) c1 e+c2 e e,
proving that z 2 BHe . At the same time, since c1 ^ c2 = 0, it follows that c1 c2 = 0.
This implies that (c1 jf (x)j) (c2 jf (y)j) = 0, yielding that (c1 jf (x)j) ^ (c2 jf (y)j) = 0.
By [4, Theorem 8.12] and since f is A-linear, we can conclude that
jf (z)j = jf (c1 x + c2 y)j = jc1 f (x) + c2 f (y)j = c1 jf (x)j+c2 jf (y)j = jf (x)j_jf (y)j :
Thus, for each x; y 2 BHe there exists z 2 BHe such that z
x and z
y. It
follows that (BHe ; ) is a directed set and fjf (x)jgx2BHe is an increasing net. By
[6, Theorem 7.14 and Theorem 7.50], we can conclude that fjf (x)jgx2BHe is order
bounded. Therefore, there exists c 2 A+ such that jf (x)j c for all x 2 BHe . Next,
1
and xn = an x for all n 2 N. It is
consider x 2 He . De…ne an = N (x) + n1 e
immediate to see that xn 2 BHe . By the previous part of the proof, this implies that
jf (an x)j c for all n 2 N. Since f is A-linear, we can conclude that
jf (x)j
c N (x) +
1
e
n
8n 2 N:
By taking the limit and since the topology induced by d is solid and x was arbitrarily
chosen, it follows that
jf (x)j cN (x)
8x 2 H;
(12)
proving the statement.
2. For each a; b 2 A and for each y; z 2 H
S (a y + b z) (x) = hx; a y + b ziH = a hx; yiH + b hx; ziH
= aS (y) (x) + bS (z) (x)
8x 2 H;
= (a S (y)) (x) + (b S (z)) (x)
in other words, we have that S (a y + b z) = a S (y) + b S (z), that is, S is a
module homomorphism. We next show that S is an isometry. Consider z 2 H. By
10
Consider b1 ; b2 2 A+ . By [11, Lemma 3], there exists c1 2 A such that 0
+
and c1 (b1 b2 ) = (b1 b2 ) . De…ne c2 = e c1 . It follows that
c1 b1 + c2 b2 = b2 + c1 (b1
b2 ) = b2 + (b1
It is enough to set b1 = jf (x)j and b2 = jf (y)j.
25
+
b2 ) = b2 + (b1
c1
e, c1 ^(e
b 2 ) _ 0 = b1 _ b 2 :
c1 ) = 0,
Proposition 4 (see also Remark 2), we have that for each x 2 H and for each n 2 N
1
jS (z) (x)j = jhx; ziH j
1
hx; xiH2 hz; ziH2 = N (x) N (z)
N (x) +
1
e N (z) :
n
This implies that
jS (z) (x)j
N (x) + n1 e
8x 2 H; 8n 2 N;
N (z)
which yields that
N (S (z)) = sup sup
x2H
n2N
jS (z) (x)j
N (x) + n1 e
N (z) :
Consider now x = z. It follows that
sup
n2N
N 2 (z)
jS (z) (z)j
=
sup
= N (z) ; 11
N (z) + n1 e n2N N (z) + n1 e
yielding that
N (S (z)) = sup sup
x2H
n2N
jS (z) (x)j
N (x) + n1 e
sup
n2N
jS (z) (z)j
= N (z)
N (z) + n1 e
and proving that N (S (z)) = N (z). Since z was arbitrarily chosen, we have that for
each x; y 2 H
dH (S (x) ; S (y)) = d (0; N (S (x)
= d (0; N (x
S (y))) = d (0; N (S (x
y)))
y)) = dH (x; y) ;
proving the statement.
3. If H is self-dual, it is immediate to see that S is onto. As for dH completeness
of H, we postpone the proof. It will be the implication "(ii) implies (i)" in Theorem
5.
Consider a 2 A+ . De…ne an = a + n1 e for all n 2 N. It follows that an 1 n2N is a well de…ned
and increasing sequence, and so is an 1 a2 n2N . Note that a an , thus aan 1 e for all n 2 N. Note
also that
1
1
a2 + a = a a + e = aan 8n 2 N;
n
n
11
that is,
a
an 1 a2 =
1
aa 1
n n
1
e 8n 2 N:
n
It follows that
d an 1 a2 ; a = ' a
an 1 a2 ^ e
'
1
e
n
! 0:
By [4, Theorem 8.43] and since the topology generated by d is locally solid and Hausdor¤, we can
conclude that supn an 1 a2 = a. It is enough to set a = N (z).
26
4.3
Dual module as a congruent space
In this subsection, we consider an Arens algebra of L1 type. We show that the dual
module unit ball, BH , is compact with respect to a weak topology, thus providing a
sort of Banach-Alaoglu theorem for the dual module. As a corollary, we obtain that
H can be seen as the norm dual of a speci…c Banach space.
Fix x 2 H. De…ne ` : H ! R by
8f 2 H :
` (f ) = ' (f (x))
Note that for each f; g 2 H and ;
(13)
2R
` ( f + g) = ' (( f + g) (x)) = ' ( f (x) + g (x))
= ' (f (x)) + ' (g (x)) = ` (f ) + ` (g) ;
that is, ` is linear. Similarly, we have that
j` (f )j = j' (f (x))j
k'kA kf (x)kA
k'kA kxkH kf kH
8f 2 H ;
that is, ` 2 (H ) where (H ) is the k kH norm dual of H . De…ne
V = f` 2 (H ) : ` is de…ned as in (13)g
and V as the k k(H
)
norm closure of V in (H ) .
Theorem 1 If A is an Arens algebra of L1 type and H a pre-Hilbert A-module, then
BH is (H ; V ) compact.
Proof. By de…nition of V , note that we need to show that for each net ffi gi2I BH
BH such that ` fij ! ` (f ) for all
there exists f 2 BH and a subnet fij j2J
` 2 V . This is equivalent to show that
' fij (x) ! ' (f (x))
8x 2 H:
Before starting, recall that k k1 : A ! [0; 1) is de…ned by kak1 = ' (jaj) for all
a 2 A. Recall also that A0 denotes the k k1 dual of A. It follows that hA; A0 i is a
Riesz dual system. Since ' is order continuous, it is immediate to verify that k k1
is indeed a well de…ned order continuous norm. This implies that (A; A0 ) is order
continuous. Moreover, by [6, Theorem 3.57] and since A is Dedekind complete, we
have that order intervals are (A; A0 ) compact. We conclude the …rst part of the proof
by …xing b 2 A and proving that the map L : A ! A, de…ned by L (a) = ba for all
a 2 A, is (A; A0 )
(A; A0 ) continuous. It is immediate to check that L is linear.
Next, we show that L is norm bounded. In fact, kL (a)k1 = ' (jbaj) = ' (jbj jaj)
27
' (kbkA e jaj) kbkA ' (jaj)
conclude that L is (A; A0 )
kbkA kak1 for all a 2 A. By [4, Theorem 6.17], we can
(A; A0 ) continuous.
(A; A0 ) topology and AH with the
In this proof, we consider A endowed with the
product topology.
Consider a net ffi gi2I
AH . For each x 2 H de…ne
BH . It follows that ffi gi2I
Ax = fa 2 A : 9i 2 I s.t. fi (x) = ag :
Note that kfi (x)kA kfi kH kxkH for all i 2 I and for all x 2 H. Since ffi gi2I
we can conclude that
Ax
[ kxkH e; kxkH e]
BH ,
8x 2 H:
Thus, we have that
ffi gi2I
x2H Ax
[ kxkH e; kxkH e]
x2H
AH
where the last by one set is compact by Tychono¤’s theorem. We can thus extract a
(A;A0 )
subnet fij j2J ffi gi2I such that fij (x) ! ax 2 [ kxkH e; kxkH e] for all x 2 H.
De…ne f : H ! A by f (x) = ax for all x 2 H. Note that f is well de…ned and A-linear.
For, consider a; b 2 A and y; z 2 H
fij (y)
(A;A0 )
!
f (y) ; fij (z)
(A;A0 )
!
f (z) =) afij (y)
(A;A0 )
!
af (y) ; bfij (z)
=) afij (y) + bfij (z)
(A;A0 )
!
(A;A0 )
!
bf (z)
af (y) + bf (z) :
(A;A0 )
At the same time, afij (y) + bfij (z) = fij (a y + b z) ! f (a y + b z). By the
uniqueness of the limit, we can conclude that f (a y + b z) = af (y) + bf (z). Since
a; b 2 A and y; z 2 H were arbitrarily chosen, it follows that f is A-linear. In particular,
we have that f is linear. We next show that f is norm bounded. In fact, recall that
f (x) 2 [ kxkH e; kxkH e] for all x 2 H. This implies that
kf (x)kA
kxkH
8x 2 H;
that is, f is k kH k kA continuous and kf kH
' 2 A0 , we can conclude that fij j2J ffi gi2I
(14)
1. Thus, f belongs to BH . Since
' fij (x) ! ' (f (x))
8x 2 H
and f 2 BH .
Next, we show that the norm dual of V; k k(H
J : H ! V by f 7! J (f ) where
)
, V , is congruent to H . De…ne
8f 2 H ; 8` 2 V:
J (f ) (`) = ` (f )
28
Lemma 4 If A is an Arens algebra of L1 type and H a pre-Hilbert A-module, then J
is a well de…ned onto linear isometry. Moreover, J is (H ; V )
(V ; V ) continuous.
Proof. The statement follows by replicating the arguments contained in [32] (see also
[24] or [21, p. 211]).
5
5.1
Self-duality
Arens algebras of L1 type
Using the results derived in Section 4, we start by observing that a necessary condition
for self-duality is the compactness of BH in the (H; S (H)) topology. In what follows,
we provide the results that will be instrumental in showing that such a condition is
also su¢ cient. We then move to state our …rst result on self-duality (Theorem 3). We
conclude by giving a su¢ cient condition and a di¤erent necessary one for self-duality
(Propositions 16 and 17).
Proposition 14 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module.
If H is self-dual, then BH is (H; S (H)) compact.
Proof. Since S is an isometry and H is self-dual, it follows that S (BH ) = BH .
Consider a net fyi gi2I BH and de…ne ffi gi2I BH to be such that fi = S (yi ) for
all i 2 I. By Theorem 1, we have that BH is (H ; V ) compact. This implies that
there exists a subnet fij j2J
ffi gi2I and f 2 BH such that ` fij ! ` (f ) for
all ` 2 V , that is, ' fij (x) ! ' (f (x)) for all x 2 H. Since S (BH ) = BH ,
there exists y 2 BH such that f = S (y). We can conclude that ' yij ; x H =
' x; yij H = ' fij (x) ! ' (f (x)) = ' (hx; yiH ) = ' (hy; xiH ) for all x 2 H, that
is, yij
(H;S(H))
!
y 2 BH .
Corollary 4 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module. If
H is self-dual, then BH is H; clk kH (S (H)) compact. In particular, BH is (H; H 0 )
compact.
Proof. Consider fyi gi2I
BH . By Proposition 14, there exists yij
(H;S(H))
j2J
fyi gi2I
and y 2 BH such that yij
!
y. Since BH is k kH bounded, this implies that
l yij ! l (y) for all l 2 clk kH (S (H)). By Lemma 1 and since H is a pre-Hilbert
space, clk kH 0 (S (H)) = H 0 . By (10), we can conclude that H 0 clk kH (S (H)), proving
(H; H 0 ) compactness.
Proposition 15 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module.
If BH is (H; S (H)) compact, then S (H)\BH is (H ; V ) and H ; V compact.
29
Proof. Consider a net ffi gi2I
S (H) \ BH . By assumption and since S is an
isometry, it follows that there exists fyi gi2I BH such that S (yi ) = fi for all i 2 I.
Since BH is (H; S (H)) compact, we have that there exists a subnet yij j2J and
y 2 BH such that
' yij ; z H ! ' (hy; ziH )
8z 2 H:
De…ne f = h ; yiH 2 S (H) \ BH . We can conclude that
S (H) \ BH and f 2 S (H) \ BH are such that
' fij (z) ! ' (f (z))
proving that S (H) \ BH
is
V such that `n
j` (f )
`n (f )j + `n (f )
` fij
k`
2 k`
`n k(H
ffi gi2I
j2J
8z 2 H;
(H ; V ) compact. Next, consider ` 2 V . It follows
that there exists f`n gn2N
` (f )
fij
)
`n k(H
k k(H
!
)
`n fij
kf kH + `n (f )
)
`. Note that
+ `n (f )
+ `n fij
`n fij
`n fij
` fij
+ k`
`n k(H
)
fij
H
:
By taking the limit in j, we have that
0
lim sup ` (f )
j
` fij
2 k`
`n k(H
8n 2 N:
)
By taking the limit in n, we can conclude that
0
lim inf ` (f )
j
` fij
lim sup ` (f )
j
` fij
proving that ` fij ! ` (f ). It follows that S (H) \ BH is
0;
H ;V
compact.
Corollary 5 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module. If
BH is (H; S (H)) compact, then S (H) is
H ; V closed.
Proof. By Lemma 4, (H ; k kH ) can be identi…ed with the dual of V; k k(H
)
. It is
then immediate to see that (H ; k kH ) can be identi…ed with the dual of V ; k k(H
)
.
By Proposition 15, S (H) \ BH is H ; V closed. By the Krein-Smulian Theorem
(see [31, Corollary 2.7.12]) and since S (H) is a vector subspace of H , it follows that
S (H) is
H ; V closed.
Theorem 2 If A is an Arens algebra of L1 type and H a pre-Hilbert A-module, then
S (H) separates the points of V .
30
Proof. We proceed by steps. Before starting, we denote by Hp the norm completion
of H with respect to k kp and A1 the norm completion of A with respect to k k1 (see
also Remark 4). We still denote by ' the extension of ' from A to A1 .
Step 1. If f is an A-linear and k kH k kA continuous operator, then f admits a
unique k kp k k1 continuous linear extension f : Hp ! A1 .
Proof of the Step. By Proposition 8 and since f is linear, the statement trivially follows.
Step 2. Let f`n gn2N
f 2 H . If `n
k k(H
!
)
V and fxn gn2N
H such that `n (f ) = ' (f (xn )) for all
k kp
` 2 (H ) , then xn ! x 2 Hp . Moreover, we have that
` (f ) = ' f (x)
8f 2 H :
Proof of the Step. Consider x 2 H. De…ne `x : H ! R by `x (f ) = ' (f (x)) for all
f 2 H . By de…nition of k k(H ) and the proof of Proposition 10, it follows that
k`x k(H
Since `n
)
= sup j' (f (x))j
sup
f 2BH
k k(H
!
)
f 2BH \S (H)
`, then fxn gn2N
j' (f (x))j = sup j' (hx; yiH )j = kxkp :
y2BH
H is a k kp Cauchy sequence which thus converges
to an element in the completion. Finally, …x f 2 H . Since `n
k k(H
!
)
`, it follows that
' (f (xn )) = `n (f ) ! ` (f ) :
k kp
kk
At the same time, by Step 1 and since xn ! x 2 Hp , we have that f (xn ) = f (xn ) !1
f (x). This implies that
`n (f ) = ' (f (xn )) ! ' f (x) :
Since f was arbitrarily chosen, we can conclude that ` (f ) = ' f (x) for all f 2 H .
Step 3. For each y 2 H and x 2 Hp
' S (y) (x) = hy; xim :
Proof of the Step. Consider y 2 H and x 2 Hp . There exists fxn gn2N
H such that
k kp
xn ! x. It follows that
kk
S (y) (xn ) = S (y) (xn ) !1 S (y) (x) :
This implies that
hy; xn im = hxn ; yim = ' (hxn ; yiH ) = ' (S (y) (xn )) ! ' S (y) (x) :
31
At the same time, by Lemma 2, hy; xn im = hy; xn im ! hy; xim , proving the statement.
We are ready to prove the main statement. Consider `; `0 2 V . By Step 2, it follows
that there exists x; x0 2 Hp such that
` (f ) = ' f (x) and `0 (f ) = ' f (x0 )
8f 2 H :
(15)
Assume that
` (f ) = `0 (f )
8f 2 S (H) :
It follows that
' S (y) (x) = ' S (y) (x0 )
8y 2 H:
By Step 3, this implies that
hy; xim = hy; x0 im
8y 2 H:
By Lemma 2, x = x0 . By (15), this proves that ` = `0 .
We are now ready to state our …rst result on self-dual modules.
Theorem 3 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module.
The following statements are equivalent:
(i) H is self-dual;
(ii) BH is
(iii) BH is
(iv) H is
(H; S (H)) compact;
H; clk kH (S (H)) compact;
H; clk kH (S (H)) sequentially complete;
(v) BH is k km complete.
Proof. Before starting recall that H 0 (resp., H ) denotes the norm dual of H when
H is endowed with the norm k km (resp., k kH ). We denote by H 00 the second dual
of H (that is, the norm dual of H 0 when H 0 is endowed with the norm k kH 0 ). Given
Corollary 2, it is well known that H 0 is an Hilbert space (see [9] or [10, Exercise pp.
149-150]). By Lemma 1, we have that S : H ! H 0 is a k km k kH 0 linear isometry.
(i) implies (ii). It is Proposition 14. (ii) implies (iii). It is Corollary 4.
(iii) implies (i). Since S (H)
clk kH (S (H)), BH is
(H; S (H)) compact. Recall
that (H ; k kH ) can be identi…ed with the norm dual of V ; k k(H
)
and S (H)
is a vector subspace of H . By Theorem 2, S (H) separates the points of V . By [4,
32
Corollary 5.108], it follows that S (H) is
H ;V
S (H) is
H ; V closed, proving S (H) = H .
dense in H . By Corollary 5,
Thus, we have showed that (i), (ii), and (iii) are equivalent.
(iii) implies (iv). By point 4. of Proposition 12 and since H is self-dual, (H; k kH )
is a Banach space. By [24], we have that the Banach space clk kH (S (H)) ; k kH
has a dual that can be identi…ed with (H; k kH ). If we consider a H; clk kH (S (H))
Cauchy sequence in H, then it is a weakly* Cauchy sequence. By [31, Corollary 2.6.21],
we can conclude that H is
H; clk kH (S (H)) sequentially complete.
(iv) implies (v). Consider a k km Cauchy sequence fxn gn2N
jl (xn )
klkH 0 kxn
l (xk )j
BH . It follows that
8n; k 2 N; 8l 2 H 0 :
x k km
(16)
R is a Cauchy sequence for each
Since S (H)
H 0 , this implies that fl (xn )gn2N
l 2 S (H). Next, consider l 2 clk kH (S (H)). Consider " > 0. It follows that there
exists l 2 S (H) such that l l H < "=4. Since fxn gn2N BH , this implies that for
each n; k 2 N
l (xn
xk )
l (xn
xk )
l
l
H
kxn
x k kH
2 l
l
H
"
:
2
(17)
At the same time, since fl (xn )gn2N
R is a Cauchy sequence, there exists n";l 2 N
such that
"
8n; k n";l :
(18)
jl (xn ) l (xk )j <
4
By (17) and (18), we can conclude that
l (xn
xk )
l (xn
xk )
l (xn
xk ) +jl (xn )
" "
+ <"
2 4
l (xk )j
8n; k
n";l :
Since " was arbitrarily chosen, it follows that l (xn ) n2N R is a Cauchy sequence.
Since l was arbitrarily chosen, it follows that fl (xn )gn2N R is a Cauchy sequence for
all l 2 clk kH (S (H)). Since H is
H; clk kH (S (H)) sequentially complete, we have
that there exists x 2 H such that l (xn ) ! l (x) for all l 2 clk kH (S (H)). By (16) and
since fxn gn2N is a k km Cauchy sequence, we also have that for each " > 0 and for each
l 2 BH 0 there exists n" 2 N such that
jl (xn )
l (xk )j <
"
2
8n; k
n" :
By taking the limit in k and since H 0 clk kH (S (H)), this implies that jl (xn x)j =
jl (xn ) l (x)j "=2 for all l 2 BH 0 and for all n n" . We thus have that kx xn km =
supl2BH 0 jl (xn x)j < " for all n
that x 2 BH . Note that
' (jN (x)
N (xn )j)
kk
n" . It follows that xn !m x. We are left to show
' (jN (x
xn )j) = kx
33
x n kp
kx
xn km ! 0;
kk
proving that N (xn ) !1 N (x). Since N (xn ) e for all n 2 N and the topology induced
by k k1 is locally solid and Hausdor¤, it follows that N (x) e, proving that x 2 BH .
(v) implies (ii). Since S is a linear isometry, if BH is k km complete, then S (BH ) is
convex and k kH 0 closed. It follows that S (BH ) is (H 0 ; H 00 ) closed. Moreover, since
BH
Bm , S (BH ) turns out to be contained in the unit ball induced by k kH 0 . Since
0
H is an Hilbert space (thus, re‡exive), this latter set is (H 0 ; H 00 ) compact. We can
conclude that S (BH ) is (H 0 ; H 00 ) compact. By de…nition of S, it is then immediate
to see that this yields the desired compactness of BH in the (H; S (H)) topology.
We conclude by providing a su¢ cient condition for the self-duality of H (Proposition
16) and a necessary one (Proposition 17).
Proposition 16 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module.
If BH is dH complete, then H is self-dual.
Proof. By Theorem 3, it is su¢ cient to show that BH is k km complete. Consider a
k km Cauchy sequence fxn gn2N BH . By (8) and (9), it follows that fxn gn2N is a dH
Cauchy sequence. Since BH is dH complete, it follows that there exists x 2 BH such that
dH
xn !
x. De…ne fyn gn2N by yn = 21 xn for all n 2 N and y = 21 x. Since fxn gn2N BH
and x 2 BH , it follows that 0
note that
0
kyn
yk2m = ' N (yn
= ' (N (yn
N (yn
y)
d
H
e for all n 2 N. Moreover, yn !
y. Next,
y)2 = ' (N (yn
y) N (yn
y))
' (N (yn
y))
y) ^ e) = dH (yn ; y) ! 0:
kk
By (5), this implies that 0 = limn 2 kyn ykm = limn kxn xkm , proving that xn !m x
and that BH is k km complete.
As mentioned in the introduction, Kaplansky [25] …rst studied self-duality for preHilbert A-modules H where A was a commutative AW -algebra.12 In such a context,
he proved that self-duality is equivalent to H satisfying certain algebraic property which
we summarize in the next de…nition (see also [16]). Proposition 17 shows that these
algebraic conditions are necessary for self-duality also in the real case. It has eluded
us whether they are also su¢ cient.
De…nition 9 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module.
We say that
1. fei gi2I is an orthogonal partition of the unit e if and only if e = supi2I ei and
ei ^ ej = 0 for all i 6= j.
12
Recall that the the main distinction with the current work is the focus on complex valued algebras
A.
34
2. H is an Hilbert-Kaplansky module if and only if for each orthogonal partition of
the unit fei gi2I A
(a) for each x 2 H, ei x = 0 for all i 2 I implies x = 0;
(b) for each k kH bounded collection fxi gi2I
ei xi = ei x.
H there exists x 2 H such that
Proposition 17 Let A be an Arens algebra of L1 type and H a pre-Hilbert A-module.
If H is self-dual, then H is an Hilbert-Kaplansky module.
Proof. Consider fei gi2I
A+ such that kei kA
1 for all i 2 I and such that
0
supi2I ei = e and ei ^ ei0 = 0 for all i 6= i . Note that ei ei0 = 0 for all i 6= i0 . Moreover,
P
we can also conclude that tB = i2B ei = _i2B ei " e where B is a …nite subset of I and
this …nite subsets are directed by using the inclusion relation. This latter observation
yields that ei ei = ei for all i 2 I.13 Finally, since ' is order continuous, observe that
0 ' (e tB ) # 0.
We start by providing a construction which will be instrumental in proving the
statement. Assume that fxi gi2I
H is k kH bounded. Call 2I0 the collection of all
…nite subsets of I. Direct 2I0 with the inclusion relation. De…ne fsB gB22I H by
0
X
sB =
ei xi
8B 2 2I0 :
i2B
P
Note that hsB ; sB iH = i2B ei hxi ; xi iH for all B.14 Since fxi gi2I H is k kH bounded,
we have that there exists M > 0 such that kxi kH M for all i 2 I. By Proposition 5
13
Note that for each i 2 I and B
jei
ei tB j = jei e
ei tB j = jei j je
tB j = ei (e
tB )
(e
tB ) :
By passing to the order limit, we have that ei tB " ei for all i 2 I. Since ei tB = ei ei for all B such
that fig B, the statement follows.
14
First, consider B and i0 62 B. Observe that
ei0 hsB ; xi0 iH = ei0
= ei0
*
X
ei xi ; xi0
i2B
X
i2B
+
H
ei hxi ; xi0 iH =
= ei0
X
i2B
X
i2B
hei xi ; xi0 iH
(ei0 ei ) hxi ; xi0 iH = 0:
Thus, we have that:
- if jBj = 1, then
hsB ; sB iH = hei xi ; ei xi iH = e2i hxi ; xi iH = ei hxi ; xi iH ;
- if the statement is true for jB 0 j = m, then consider B such that jBj = m + 1. It follows that
B = B 0 [ fi0 g :
35
and since each ' in K is an algebra homomorphism, it follows that for each ' 2 K
!
X
X
' (ei hxi ; xi iH )
0 ' (hsB ; sB iH ) = '
ei hxi ; xi iH =
=
X
i2B
i2B
' (ei ) ' (hxi ; xi iH )
= M 2'
X
ei
i2B
!
X
i2B
i2B
' (ei ) M 2
i2B
8B 2 2I0 :
M 2 ' (e) = M 2
By Proposition 5, this implies that
q
ksB kH = max ' (hsB ; sB iH )
8i 2 I:
M
'2K
Since BH is
X
' (ei ) kxi k2H
(H; S (H)) compact, it follows that there exist a subnet sBj
x 2 H such that sBj
(H;S(H))
!
'
j2J
and
x 2 H, that is,
sB j ; y
H
! ' (hx; yiH )
8y 2 H:
1. Consider xi = x for all i 2 I. It is immediate to see that fxi gi2I is k kH bounded.
Next, we show that x = x. Note that for each j 2 J and for each y 2 H
' (hx; yiH )
'
sBj ; y
H
= '
e x
0*0
= ' @ @e
= '
e
sBj ; y
X
i2Bj
H
1
ei A x; y
= khx; yiH kA ' e
H
A
' hx; yiH e
tBj hx; yiH
= ' jhx; yiH j e
+ 1
tBj
' khx; yiH kA e e
tBj
tBj
tBj :
By passing to the limit with respect to j, we obtain that ' sBj ; y H ! ' (hx; yiH )
for all y 2 H. By the uniqueness of the limit, we obtain that x = x. Thus, if ei x = 0
for all i 2 I, then sBj = 0 for all j 2 J, implying that x = x = 0.
We thus have that
hsB ; sB iH = hsB 0 + ei0 xi0 ; sB 0 + ei0 xi0 iH
= hsB 0 ; sB 0 + ei0 xi0 iH + hei0 xi0 ; sB 0 + ei0 xi0 iH
= hsB 0 ; sB 0 iH + 2ei0 hsB 0 ; xi0 iH + hei0 xi0 ; ei0 xi0 iH
= hsB 0 ; sB 0 iH + ei0 hxi0 ; xi0 iH
X
X
=
ei hxi ; xi iH + ei0 hxi0 ; xi0 iH =
ei hxi ; xi iH ;
i2B 0
i2B
proving that the statement is true for m + 1. The whole statement follows by induction.
36
2. Assume that fxi gi2I H is k kH bounded. Consider { 2 I. Consider j 2 J such
that Bj f{g. It follows that
0
1 0
1
X
X
e{ sBj = e{ @
ei xi A = @
(e{ ei ) xi A = e{ x{ :
i2Bj
i2Bj
We can conclude that, for each y 2 H, eventually
'
e{ sBj ; y
H
= ' (he{ x{ ; yiH )
thus, proving that
(H;S(H))
!
e{ sBj
At the same time, since '
each y 2 H
lim '
j
e{ sBj ; y
H
sBj ; y
= lim '
j
H
e{ x{ :
! ' (hx; yiH ) for all y 2 H, we have that for
sBj ; e{ y
H
= ' (hx; e{ yiH ) = ' (he{ x; yiH ) ;
(H;S(H))
that is, e{ sBj
!
e{ x. By the uniqueness of the limit, we can conclude that
e{ x{ = e{ x. Since { was arbitrarily chosen, the statement follows.
5.2
Finite dimensional case
In this subsection, we discuss separately the case in which A is a …nite dimensional
Arens algebra, that is, A is isomorphic to some Rn . It is immediate to see that if A is
…nite dimensional, then it is of L1 type. At the same time, the …nite dimensional case
merits to be discussed separately. First, the result of self-duality can be obtained via
more direct methods. Second, it is the only case in which we can show that Hilbert
A-modules are indeed self-dual. In other words, we can show that k kH completeness
is necessary and su¢ cient for self-duality.
Theorem 4 Let A be a …nite dimensional Arens algebra and H a pre-Hilbert Amodule. The following statements are equivalent:
(i) H is k kH complete, that is, H is an Hilbert A-module;
(ii) H is k km complete;
(iii) H is self-dual.
Proof. Before starting, observe that, since A is …nite dimensional, it is Dedekind
complete and admits a strictly positive functional ' (see Proposition 9).
37
(i) implies (ii). By Proposition 9 and since A is …nite dimensional, k km and k kH
are equivalent. It follows that H is k km complete.
(ii) implies (iii). By Corollary 2 and since H is k km complete, it follows that H
is an Hilbert space with inner product h ; im . Consider f : H ! A which is A-linear
and bounded. By Proposition 8, it follows that f : H ! A is A-linear and k km k k1
continuous. Consider the linear functional l = ' f . Since ' is k k1 continuous and f
is k km k k1 continuous, we have that l is k km continuous. By the standard Riesz
representation theorem, there exists (a unique) y 2 H such that l (x) = hx; yim for all
x 2 H. It follows that
' (f (x)
hx; yiH ) = ' (f (x))
Fix x 2 H. De…ne a = f (x)
hx; yim = 0
8x 2 H: (19)
hx; yiH 2 A. By (19), we have that
0 = ' (f (a x)
= ' (a (f (x)
' (hx; yiH ) = l (x)
ha x; yiH ) = ' (af (x)
a hx; yiH )
hx; yiH )) = ' (aa) = ' a2 :
Since ' is strictly positive, this implies that a2 = 0. Since A is an Arens algebra, we
can conclude that f (x) hx; yiH = a = 0. Since x was arbitrarily chosen, it follows
that f (x) = hx; yiH for all x 2 H, proving that H is self-dual.
(iii) implies (i). By point 4. of Proposition 12, it follows that H is k kH complete.
5.3
f -algebras of L0 type
If A is an f -algebra of L0 type and H is a pre-Hilbert A-module, then the inner product
h ; iH still satis…es the Cauchy-Schwarz inequality as in points 1. and 2. of Proposition
4 (see also Remark 2). By Proposition 7, we can endow H with the invariant metric
dH . We de…ne
n
o
1
He = x 2 H : hx; xiH2 2 Ae :
Finally, we de…ne
e
H e = f 2 AH
e : f is Ae -linear and bounded :
Observe that f 2 He only if there exists 0
all x 2 He .
c 2 Ae such that f 2 (x)
c hx; xiH for
Proposition 18 Let A be an f -algebra of L0 type and H a pre-Hilbert A-module. The
following statements are true:
1. Ae is an Arens algebra of L1 type;
38
2. He is a pre-Hilbert Ae -module;
3. He is dH dense in H;
4. He is the dual module of He .
Proof. 1. and 2. By de…nition, Ae is an Arens algebra of L1 type. Next, we show
that He is closed under +. Consider x; y 2 He . By the Cauchy-Schwarz inequality, we
have that
hx + y; x + yiH = hx; xiH + 2 hx; yiH + hy; yiH
1
1
hx; xiH + 2 hx; xiH2 hy; yiH2 + hy; yiH 2 Ae :
1
Next, we show that if a 2 Ae and x 2 He , then a x 2 He . Since a; hx; xiH2 2 Ae and
Ae is an Arens algebra, it follows that ha x; a xiH = a2 hx; xiH 2 Ae , proving that
1
ha x; a xiH2 2 Ae . The closure of He with respect to + and yields that (He ; +) is an
abelian (sub)group and properties (1)-(4) of De…nition 1 automatically follow. Finally,
1
1
1
1
note that if x; y 2 He , then hx; xiH2 ; hy; yiH2 2 Ae and jhx; yiH j hx; xiH2 hy; yiH2 2 Ae ,
proving that hx; yiH 2 Ae . Thus, h ; iHe is h ; iH restricted to He He . Then,
properties (5)-(8) of De…nition 1 automatically follow.
3. Since A is an f -algebra of L0 type, it follows that for each c 2 A+ there exists
fen gn2N
Ae such that 0
en " e, e2n = en , and en c 2 Ae for all n 2 N. Consider
1
c = hx; xiH2 = N (x) for some x 2 H. De…ne xn = en x for all n 2 N. It follows that
N (xn ) = en N (x) 2 Ae , that is, xn 2 He . Moreover, note that ((e en ) N (x)) ^ e # 0.
Since ' is order continuous, it follows that
xn ) ^ e) = ' (N (e x
dH (x; xn ) = ' (N (x
= ' (((e
en x) ^ e) = ' (N ((e
en ) x) ^ e)
en ) N (x)) ^ e) ! 0;
d
H
proving that xn !
x. Since x was arbitrarily chosen, it follows that He is dH dense in
H.
4. It follows by de…nition of dual module.
Our proof of self-duality for a pre-Hilbert A-module H where A is an f -algebra of
0
L type hinges on our self-duality result for a pre-Hilbert A-module H where A is an
algebra of L1 type. This latter pre-Hilbert module will be He and the algebra will be
the subalgebra Ae .
Theorem 5 Let A be an f -algebra of L0 type and H a pre-Hilbert A-module. The
following statements are equivalent:
(i) H is dH complete;
39
(ii) H is self-dual.
Proof. (i) implies (ii). By Proposition 18, we have that He is a pre-Hilbert Ae -module
and Ae is an Arens algebra of L1 type. We can thus de…ne k kHe and BHe . It follows
that x 2 BHe if and only if hx; xiH e. We proceed by steps.
Step 1. If f 2 H , then f is dH
d continuous.
Proof of the Step. By point 1. of Proposition 13, the statement follows.
Step 2. BHe is dH complete.
Proof of the Step. Consider a dH Cauchy sequence fxn gn2N
BHe . Since H is dH
dH
complete, it follows that there exists x 2 H such that xn ! x. We next show that x 2
BHe . By one of Birkho¤’s inequalities (see [6, Theorem 1.9]) and since hxn ; xn iH
e
for all n 2 N, we also have that
je
N (x) _ ej = jN (xn ) _ e
N (x) _ ej
jN (xn )
N (x)j
N (xn
x)
8n 2 N:
It follows that
je
N (x) _ ej ^ e
N (xn
x) ^ e
8n 2 N:
We have that
0
' (je
N (x) _ ej ^ e)
' (N (xn
x) ^ e) = dH (xn ; x) ! 0:
This implies that
' (je
N (x) _ ej ^ e) = 0 =) je
N (x) _ ej ^ e = 0 =) je
=) e = N (x) _ e =) N (x)
that is, hx; xiH
N (x) _ ej = 0
e;
e and x 2 BHe .
Step 3. The pre-Hilbert Ae -module He is self-dual, that is, for each f 2 He there exists
z 2 He such that f (x) = hx; ziH for all x 2 He .
Proof of the Step. By points 1. and 2. of Proposition 18, Step 2, and Proposition 16,
the statement follows.
Step 4. For each f 2 H there exists z 2 H such that f (x) = hx; ziH for all x 2 He .
Proof of the Step. Consider f 2 H . It follows that there exists c 2 A+ such that
f 2 (x)
c hx; xiH for all x 2 H. Since A is an f -algebra of L0 type, it follows that
there exists a sequence fen gn2N Ae such that 0 en " e, e2n = en , and en c 2 Ae for
all n 2 N. De…ne fn = en f for all n 2 N. Note that for each x 2 H
d (fn (x) ; f (x)) = ' (jf (x)
fn (x)j ^ e) = ' (((e
40
en ) jf (x)j) ^ e) ! 0:
(20)
Fix n 2 N. By Proposition 11, we have that fn is A-linear. In particular, fn2 (x)
e2n c hx; xiH = cn hx; xiH for all x 2 H where cn = en c 2 Ae . Thus, fn restricted to He
belongs to He . By Step 3, there exists a unique zn 2 He such that
(en f ) (x) = en f (x) = fn (x) = hx; zn iH
8x 2 He :
Note that for each n > k and for each x 2 He
((fn
fk ) (x))2 = (en f (x)
ek f (x))2 = (en
ek )2 f 2 (x)
ek )2 c hx; xiH :
(en
It follows that
hx; zn
zk iH
j(fn
fk ) (x)j
1
1
ek ) c 2 hx; xiH2
(en
8x 2 He :
(21)
Since A is an f -algebra of L0 type, we have that
Ae 3 sup hx; zn
zk iH
x2BHe
1
(en
ek ) c 2 :
By the claim contained in the proof of Proposition 10, there exists a sequence fxl gl2N
k kA
BHe such that hxl ; zn zk iH " and hxl ; zn zk iH !e hzn zk ; zn
that
1
k kA
hxl ; zn zk iH ^ e !e hzn zk ; zn zk iH2 ^ e:
1
zk iH2 . It follows
By Theorem [6, Theorem 1.8] and (21), we have that
hxl ; zn
zk iH ^ e
sup (hx; zn
x2BHe
(en
sup hx; zn
zk iH ^ e) =
1
ek ) c 2 ^ e
x2BHe
zk iH
!
^e
8l 2 N:
We can conclude that
1
dH (zn ; zk ) = ' hzn
' (en
zk ; zn
zk iH2 ^ e = lim ' (hxl ; zn
l
1
1
zk iH ^ e)
1
ek ) c 2 ^ e = d ek c 2 ; en c 2 :
It follows that fzn gn2N is a dH Cauchy sequence. Since H is dH complete, it follows
d
H
that there exists z 2 H such that zn !
z. At the same time, for each x 2 BHe we have
that
d (hx; zn iH ; hx; ziH ) = ' (jhx; ziH
'
1
2
hx; zn iH j ^ e) = ' (jhx; z
hx; xiH hz
zn ; z
= dH (z; zn ) ! 0:
41
1
2
zn iH ^ e
zn iH j ^ e)
' hz
zn ; z
1
zn iH2 ^ e
It follows that f (x) = hx; ziH for all x 2 BHe . This implies that f (x) = hx; ziH for all
x 2 He .
By Step 4, if f 2 H , then there exists z 2 H such that f (x) = hx; ziH for all
x 2 He . By Proposition 18, He is dH dense in H. By Step 1, f is dH d continuous.
This implies that f (x) = hx; ziH for all x 2 H, proving that H is self-dual.
(ii) implies (i). Consider a dH Cauchy sequence fyn gn2N . De…ne ffn gn2N as fn =
S (yn ) for all n 2 N.
d
Step 1. There exists f : He ! A such that fn (x) ! f (x) for all x 2 He .
Proof of the Step. By Proposition 4, we have that
jfn (x)
fm (x)j
N (x) N (yn
8n; m 2 N; 8x 2 H:
ym )
(22)
Fix x 2 BHe . We can conclude that
jfn (x)
fm (x)j ^ e
ym )) ^ e
(N (x) N (yn
ym ) ^ e;
N (yn
yielding that
d (fn (x) ; fm (x)) = ' (jfn (x)
' (N (yn
We thus have that ffn (x)gn2N
fm (x)j ^ e)
ym ) ^ e) = dH (yn ; ym )
8n; m 2 N:
A is a d Cauchy sequence. Since A is complete, this
d
yields that fn (x) ! ax . Next, note that if x 2 He nBHe , then x =
d
x
kxkHe
2 BHe . We
have that fn (x) ! ax . Thus, we can conclude that there exists ax 2 A such that
d
fn (x) = kxkHe fn (x) ! kxkHe ax = ax :
By the uniqueness of the limit and since x was arbitrarily chosen, we can de…ne a map
f : He ! A such that f (x) = ax for all x 2 He .
Step 2. The map f is such that
f (a x + b y) = af (x) + bf (y)
8a; b 2 Ae ; 8x; y 2 He :
Proof of the Step. Consider a; b 2 Ae and x; y 2 He . We have that a x + b y 2 He .
By Step 1 and since each fn is A-linear, this implies that
d
afn (x) + bfn (y) = fn (a x + b y) ! f (a x + b y) :
d
d
At the same time, since fn (x) ! f (x) and fn (y) ! f (y), we have that
d
afn (x) + bfn (y) ! af (x) + bf (y) :
42
By the uniqueness of the limit, we can conclude that f (a x + b y) = af (x) + bf (y),
proving the statement.
Step 3. There exists c 2 A+ such that
jf (x)j
8x 2 He :
cN (x)
(23)
In particular, f is uniformly continuous on He .
Proof of the Step. Since N is a vector-valued norm, we have that
jN (yn )
N (ym )j
It follows that fN (yn )gn2N
N (yn
8n; m 2 N:
ym )
A+ is a d Cauchy sequence. Since A is d complete, it
d
follows that there exists c 2 A+ such that N (yn ) ! c 2 A+ . By Proposition 4, we also
have that
jfn (x)j N (yn ) N (x)
8x 2 He ; 8n 2 N:
By passing to the limit and since the topology induced by d is linear and locally solid,
we can conclude that
jf (x)j cN (x)
8x 2 He ;
(24)
proving the statement. Finally, by [4, Lemma 5.17], it is enough to show continuity at
dH
d
d
0. Consider xn !
0. It follows that N (xn ) ! 0. This implies that cN (xn ) ! 0. By
(24), we have that
d (0; f (xn )) = ' (jf (xn )j ^ e)
' ((cN (xn )) ^ e) = d (0; cN (xn )) ! 0;
proving continuity at 0.
Step 4. f admits a unique extension to H which is A-linear and bounded. In particular,
there exists y 2 H such that f = S (y).
Proof of the Step. Since f is uniformly continuous and He is dH dense in H, it is well
known that f admits a unique uniform continuous extension to H. For the moment,
we denote such extension by f . Then, we will simply denote it f . Since f was additive,
it is a routine argument to show that so is f . Next, consider a 2 Ae and x 2 H. There
dH
dH
exists a sequence fxn gn2N He such that xn !
x. It follows that a xn !
a x. Since
f is a continuous extension of f and by Step 2, we can conclude that
d
af (xn ) = f (a xn ) = f (a xn ) ! f (a x) :
d
d
At the same time, we have that f (xn ) = f (xn ) ! f (x), thus, af (xn ) = af (xn ) !
af (x). By the uniqueness of the limit, we can conclude that f (a x) = af (x). Next,
d
consider a 2 A and x 2 H. There exists fan gn2N
Ae such that an ! a. We have
d
H
that an x !
a x. Since f is continuous, we can conclude that
d
an f (x) = f (an x) ! f (a x) :
43
d
At the same time, we have that an f (x) ! af (x). By the uniqueness of the limit, we can
conclude that f (a x) = af (x), proving A-linearity. Finally, consider x 2 H. There
dH
d
exists a sequence fxn gn2N He such that xn !
x. In particular, N (xn ) ! N (x). By
(23) and since f is a uniform continuous extension of f and the topology induced by d
is locally solid, we have that
f (x)
d
f (xn ) = jf (xn )j
d
cN (xn ) ! cN (x) ;
The last part of the statement follows since H is self-dual.
d
H
Step 5. fn !
f.
Proof of the Step. By (22), we have that
jfn (x) fm (x)j
N (x) + k1 e
N (yn
ym )
8k; m; n 2 N; 8x 2 H:
This yields that
d 0;
jfn (x) fm (x)j
N (x) + k1 e
dH (yn ; ym )
8k; m; n 2 N; 8x 2 H:
Consider " > 0. Since fyn gn2N is a dH Cauchy sequence, there exists n" 2 N such that
d 0;
jfn (x) fm (x)j
N (x) + k1 e
8k 2 N; 8m; n
"
n" ; 8x 2 H:
By taking the limit in n, we have that
jf (x) fm (x)j
"
8k 2 N; 8m n" ; 8x 2 H:
N (x) + k1 e
n
o
(x) fm (x)j
Next, consider the sequence jf N
^
e
. This sequence is bounded by e and
(x)+ 1 e
d 0;
k
jf (x) fm (x)j
N (x)+ k1 e
k2N
jf (x) fm (x)j
N (x)+ k1 e
^ e " supk2N
increasing. Thus,
Since ' is order continuous, we can conclude that
d 0; sup
k2N
jf (x) fm (x)j
N (x) + k1 e
^ e = supk2N
jf (x) fm (x)j
^e
N (x) + k1 e
k2N
jf (x) fm (x)j
= ' sup
^e
N (x) + k1 e
k2N
jf (x) fm (x)j
= sup '
^e
"
N (x) + k1 e
k2N
jf (x) fm (x)j
N (x)+ k1 e
^ e.
= ' sup
8m
n" ; 8x 2 H:
By the proof of point 2 of Proposition 13, we also have that
N (f
fm ) = N (S (y)
S (ym )) = N (S (y
ym )) = sup
k2N
44
jhy ym ; y ym iH j
:
N (y ym ) + k1 e
We can conclude that
dH (f; fm ) = d (0; N (f
fm ))
jf (y
= d 0; sup
k2N
ym ) fm (y ym )j
N (y ym ) + k1 e
" 8m
n" ;
proving the statement.
By point 2 of Proposition 13 and Steps 4 and 5, we have that
dH (y; yn ) = dH (f; fn ) ! 0;
proving completeness of H.
6
Examples
Consider a nonempty set , a -algebra of subsets of denoted by F, a sub- -algebra
G F, and a probability measure P : F ! [0; 1]. Two F-measurable random variables
are de…ned to be equivalent if and only if they coincide almost surely. De…ne:
1. A = L0 (G), that is, A is the space (of equivalence classes) of real valued and
G-measurable functions;15
2. b
a if and only if b (!)
a (!) almost surely;
3. e = 1 , that is, e is the function that takes constant value 1;
4. It follows that Ae = L1 (G), that is, Ae is the space of all essentially bounded
and G-measurable functions;
5. ' : L1 (G) ! R as
' (a) =
6. d : L0 (G)
Z
L0 (G) ! R as
d (a; b) = ' (jb
aj ^ e) =
adP
8a 2 L1 (G) ;
Z
aj ^ e) dP
(jb
8a; b 2 L0 (G) :
Note that the topology induced by d is the one of convergence in probability P .
It is immediate to verify that L1 (G) is an Arens algebra of L1 type and A = L0 (G)
is an f -algebra of L0 type.
15
As usual, we view the equivalence classes as functions. This convention will apply throughout the
rest of the paper.
45
6.1
The module L2;0 ( ; G; F; P )
We denote by L0 (F) = L0 ( ; F; P ) the space of real valued and F-measurable functions. We call x; y; and z the elements of L0 (F). Given an F-measurable function
x : ! R such that x 0, we denote by E (xjjG) its conditional expected value with
respect to P given G (see Loeve [27, Section 27]). Denote by
H = L2;0 ( ; G; F; P ) = x 2 L0 (F) : E x2 jjG 2 L0 (G)
n
o
p
= x 2 L0 (F) : E (x2 jjG) 2 L0 (G) :
We endow H with two operations:
1. + : H
2.
:A
H ! H which is the usual pointwise sum operation;
H ! H such that a x = ax where ax is the usual pointwise product.
The space L2;0 ( ; G; F; P ) was introduced by Hansen and Richard [19]. Finally,
we also de…ne an inner product, namely, h ; iH : H H ! L0 (G) by
hx; yiH = E (xyjjG)
8x; y 2 H:
Hansen and Richard [19, p. 592] show that h ; iH is well de…ned. They also prove the
next result.
Proposition 19 (H; +; ; h ; iH ) is a pre-Hilbert L0 (G)-module.
Note that dH : H
H!R
Z q
dH (x; y) =
E (x
y)2 jjG ^ e dP
8x; y 2 H:
Theorem 6 H is self-dual.
Proof. By Theorem 5, it is enough to check that H is dH complete. This follows from
[19, Theorem A.1].
6.2
The module L2;1 ( ; G; F; P )
We denote by L2 (F) = L2 ( ; F; P ) the space of F-measurable and square integrable
functions. Denote by
He = L2;1 ( ; G; F; P ) = x 2 L0 (F) : E x2 jjG 2 L1 (G)
n
o
p
2
1
2
= x 2 L (F) : E (x jjG) 2 L (G)
L2 (F) :
46
Since L2;1 ( ; G; F; P )
L2;0 ( ; G; F; P ), we endow He with the two operations +
and of Subsection 6.1. We also restrict (f; g) 7! E (f gjjG) to He . By Proposition 18,
it follows that L2;1 ( ; G; F; P ) is a pre-Hilbert L1 (G)-module.
Note that k kHe : He ! R is such that
q
kxkHe = kE (x2 jjG)kL1 (G)
8x 2 He :
Similarly, we have that
sZ
kxkm =
E (x2 jjG) dP =
sZ
x2 dP = kxkL2 (F )
8x 2 He :
Theorem 7 He is self-dual.
Proof. By Theorem 3, we only need to show that BHe is k km = k kL2 (F ) complete. To
this end, it is enough to show that BHe is k kL2 (F ) closed. Thus, consider fxn gn2N BHe
k kL2 (F )
such that xn ! x 2 L2 (F). This implies that there exists a subsequence fxnk gk2N
a:s
such that xnk ! x. By the conditional Fatou’s lemma (see [13, p. 340]), we have that
E x2 jjG = E lim inf x2nk jjG
lim inf E x2nk jjG
k
k
1 < 1,
proving that x 2 BHe .
6.3
The module L2;1 ( ; G; F; P ; E)
Consider an in…nite dimensional separable Hilbert space E (for example, l2 (N)).16
De…ne h ; iE to be the inner product of E. Let us denote by d and d0 generic elements
of E. Since E is separable, it admits a countable orthonormal basis: fdn gn2N . Given a
function, x : ! E we say that x is weakly measurable if and only if the real valued
function
! 7! hx (!) ; diE
is F-measurable for all d 2 E. By [7, Theorems 34.2 and 34.4], if x; y :
weakly measurable, then
! 7! hx (!) ; y (!)iE =
1
X
n=1
hx (!) ; dn iE hy (!) ; dn iE
is real valued and F-measurable. In particular, by Perseval’s identity,
! 7! hx (!) ; x (!)iE =
16
1
X
n=1
jhx (!) ; dn iE j2
A similar analysis can be carried over when E is …nite dimensional.
47
8! 2
! E are
is real valued and F-measurable. Given this observation, we denote by L2E (F) =
L2 ( ; F; P ; E) the space
Z
2
LE (F) = x 2 E : x is weakly measurable and
hx; xiE dP < 1 :
As before we identify the elements of L2E (F) whenever they coincide almost surely. At
the same time, as usual, we view the equivalence classes as functions. Thus, x 2 L2E (F)
1
if and only if x is weakly measurable and ! 7! hx (!) ; x (!)iE2 belongs to L2 (F).
Consider x; y 2 L2E (F), we showed that ! 7! hx (!) ; y (!)iE is F-measurable, we next
show it is also integrable. Since E is an Hilbert space, observe that
1
1
jhx (!) ; y (!)iE j
hx (!) ; x (!)iE2 hy (!) ; y (!)iE2
Since x; y 2 L2E (F), it follows that
Z
Z
1
1
jhx; yiE j dP
hx; xiE2 hy; yiE2 dP
sZ
hx; xiE dP
Z
8! 2 :
hy; yiE dP < 1:
(25)
(26)
In this way, for each x; y 2 L2E (F), we can de…ne the G-conditional expectation of
! 7! hx (!) ; y (!)iE , that is, E (hx; yiE jjG). We thus de…ne
q
2;1
2
H = L ( ; G; F; P ; E) = x 2 LE (F) : E (hx; xiE jjG) 2 L1 (G) :
De…ne A to be the Arens algebra L1 (G) = L1 ( ; G; P ), we endow H with two
operations:
1. + : H H ! H which is the usual pointwise sum operation, that is, (x + y) (!) =
x (!) + y (!) for all ! 2 .
2.
:A
H ! H such that (a x) (!) = a (!) x (!) for all ! 2 .17
H ! L1 (G) by
Finally, we also de…ne an inner product, namely, h ; iH : H
hx; yiH = E (hx; yiE jjG)
8 (x; y) 2 H
H:
Proposition 20 (H; +; ; h ; iH ) is a pre-Hilbert L1 (G)-module.
Proof. Consider x; y 2 H. It is straightforward to verify that x + y is weakly measurable. By (26), it follows that x + y 2 L2E (F). Similarly, by (25) and the conditional
Cauchy-Schwarz inequality, we have that
E (hx + y; x + yiE jjG) = E (hx; xiE jjG) + 2E (hx; yiE jjG) + E (hy; yiE jjG)
1
17
1
E (hx; xiE jjG) + 2E hx; xiE2 hy; yiE2 jjG + E (hy; yiE jjG)
q
E (hx; xiE jjG) + 2 E (hx; xiE jjG) E (hy; yiE jjG) + E (hy; yiE jjG) :
Observe that for each ! 2
a (!) with x (!).
, a (!) 2 R and x (!) 2 E. Thus, a (!) x (!) is the scalar product of
48
This implies that H is closed under +. It is then easy to prove that (H; +) is an abelian
group. Similarly, if a 2 A and x 2 H, then
E (ha x; a xiE jjG) = E a2 hx; xiE jjG = a2 E (hx; xiE jjG) 2 L1 (G) ;
proving that H is closed under . Properties (1)-(4) of De…nition 1 are then easily
veri…ed. This shows that H is an L1 (G)-module.
Next, by (25) and the conditional Cauchy-Schwarz inequality, we have that h ; iH
is well de…ned, and:
5. Consider x 2 H. Since hx (!) ; x (!)iE 0 for all ! 2 , we have that hx; xiH =
E (hx; xiE jjG) 0. At the same time, we can conclude that
hx; xiH = 0 () E (hx; xiE jjG) = 0 () hx; xiE = 0 () x = 0:
6. Consider x; y 2 H. We have that
hx; yiH = E (hx; yiE jjG) = E (hy; xiE jjG) = hy; xiH :
7. Consider x; y; z 2 H. We have that
hx + y; ziH = E (hx + y; ziE jjG) = E (hx; ziE + hy; ziE jjG)
= E (hx; ziE jjG) + E (hy; ziE jjG) = hx; ziH + hy; ziH :
8. Consider a 2 A and x; y 2 H. We have that
ha x; yiH = E (ha x; yiE jjG) = E (a hx; yiE jjG) = aE (hx; yiE jjG) = a hx; yiH :
We can conclude that H is a pre-Hilbert L1 (G)-module.
Note that for each x 2 H
sZ
sZ
q
kxkm = ' (hx; xiH ) =
E (hx; xiE jjG) dP =
hx; xiE dP = kxkL2 (F ) :
E
Theorem 8 If G is generated by a …nite partition where each atom has strict positive
probability, then H is self-dual.
Proof. Since G is generated by a …nite partition, L1 (G) is …nite dimensional. Moreover, we have that H = L2E (F). By Theorem 4, we only need to show that H is k km
complete. Since H = L2E (F) and k km = k kL2 (F ) , the statement follows by Diestel
E
and Uhl [12, p. 50].
49
6.3.1
A Finance illustration
N
Let
2 (0; 1] and de…ne E = fd 2 R :
hd; d iE =
1
X
t
t 2
dt
t=1
dimensional separable Hilbert space where
0
1
X
dt d0t
t=1
< 1g. The space E is an in…nite
8d; d0 2 E:
For each t 2 N, de…ne also t : E ! R such that t (d) = dt for all d 2 E.
Consider a …ltration space ; fFt gt2N ; F; P where F = ([t2N Ft ). Call G = F1 .
De…ne
H = L2;1
; fFt gt2N ; P ; E = x 2 L2;1 ( ; G; F; P ; E) :
t
x 2 L0 (Ft ) :
In words, H is the space of processes in L2;1 ( ; G; F; P ; E) that are adapted to the
…ltration. Since H is a subset of L2;1 ( ; G; F; P ; E), we endow H with the same + and
operations of Subsection 6.3. Similarly, we consider (x; y) 7! E (hx; yiE jjG) restricted
to H. It is standard to show that H is a pre-Hilbert L1 (G)-module. By Theorem 8,
it easily follows that if F1 is generated by a …nite partition where each atom has strict
positive probability, then H is self-dual.
Remark 5 In Finance, this module can be used to price in…nite streams of payo¤s.
6.4
The module M 2;1 (G)
Consider a …ltration space
; fFt gt2N ; F; P where F = ([t2N Ft ). Assume that
G is a sub- -algebra of F. We denote by M 2 the space of L2 bounded martingales.
Recall that x 2 M 2 (see, e.g., [28, p. 209]) if and only if there exists a unique (terminal
variable/…nal value) x1 2 L2 (F) such that
xt = E (x1 jjFt )
8t 2 N:
De…ne A = L1 (G) and
H = M 2;1 (G) = x 2 M 2 : E x21 jjG 2 L1 (G) :
We endow H with two operations:
1. + : H
2.
:A
H ! H which is the usual pointwise sum operation;
H ! H such that a x is de…ned by
(a x)t = E (ax1 jjFt )
We also de…ne h ; iH : H
8t 2 N:
H ! A as
hx; yiH = E (x1 y1 jjG)
50
8x; y 2 H:
(27)
Proposition 21 (H; +; ; h ; iH ) is a pre-Hilbert L1 (G)-module.
Proof. By the conditional Cauchy-Schwarz inequality, it is immediate to see that
H is closed under + and (H; +) is an abelian group. The outer product is also
well de…ned. In fact, by [28, p. 209] and (27), a x is an element of M 2 . At the
same time, since the terminal value of a x is unique, (a x)1 = ax1 2 L2 (F) and
E (a x)21 jjG = E (a2 x21 jjG) = a2 E (x21 jjG) 2 L1 (G). Observe also that:
1. Consider a 2 A, x; y 2 H, and a (x + y). It follows that
(a (x + y))t = E (a (x + y)1 jjFt ) = E (a (x1 + y1 ) jjFt )
= E (ax1 + ay1 jjFt ) = E (ax1 jjFt ) + E (ay1 jjFt )
= (a x)t + (a y)t
8t 2 N;
proving that a (x + y) = a x + a y.
2. Consider a; b 2 A, x 2 H, and (a + b) x. It follows that
((a + b) x)t = E ((a + b) x1 jjFt ) = E (ax1 + bx1 jjFt )
= E (ax1 jjFt ) + E (bx1 jjFt ) = (a x)t + (b x)t
8t 2 N;
proving that (a + b) x = a x + b x.
3. Consider a; b 2 A, x 2 H, and a (b x). It follows that
(a (b x))t = E (a (b x)1 jjFt ) = E (a (bx1 ) jjFt )
= E ((ab) x1 jjFt ) = ((ab) x)t
8t 2 N;
proving that a (b x) = (ab) x.
4. Consider e 2 A and x 2 H. It follows that
(e x)t = E (ex1 jjFt ) = E (x1 jjFt ) = xt
8t 2 N;
proving that e x = x.
This shows that H is an L1 (G)-module. Next, by the conditional Cauchy-Schwarz
inequality, we have that h ; iH is well de…ned, and:
5. Consider x 2 H. We have that hx; xiH = E (x21 jjG)
can conclude that
0. At the same time, we
hx; xiH = 0 () E x21 jjG = 0 () x1 = 0
() xt = E (x1 jjFt ) = 0 8t 2 N
() x = 0
51
6. Consider x; y 2 H. We have that
hx; yiH = E (x1 y1 jjG) = E (y1 x1 jjG) = hy; xiH :
7. Consider x; y; z 2 H. We have that
hx + y; ziH = E ((x + y)1 z1 jjG) = E ((x1 + y1 ) z1 jjG)
= E (x1 z1 + y1 z1 jjG) = hx; ziH + hy; ziH :
8. Consider a 2 A and x; y 2 H. We have that
ha x; yiH = E ((a x)1 y1 jjG) = E ((ax1 ) y1 jjG) = E (a (x1 y1 ) jjG) = a hx; yiH :
We can conclude that H is a pre-Hilbert L1 (G)-module.
Note that k kH : H ! R is such that
q
q
kxkH = khx; xiH kL1 (G) = kE (x21 jjG)kL1 (G)
8x 2 H:
Similarly, we have that
kxkm =
q
' (hx; xiH ) =
sZ
E (x21 jjG) dP = kx1 kL2 (F )
8x 2 H:
Theorem 9 H is self-dual.
Proof. By Theorem 3, we only need to show that BH is k km complete. Consider a
sequence fxn gn2N BH which is k km Cauchy. It follows that f(xn )1 gn2N is a k kL2 (F )
Cauchy sequence and
E (xn )21 jjG
8n 2 N:
e=1
(28)
k kL2 (F )
It follows that there exists an element x1 2 L2 (F) such that (xn )1 ! x1 . This
a:s
implies that there exists also a subsequence fxnk gk2N such that (xnk )1 ! x1 . By (28)
and the conditional Fatou’s lemma (see [13, p. 340]), we have that
E x21 jjG = E lim inf (xnk )21 jjG
lim inf E (xnk )21 jjG
k
k
If we de…ne x as the element in M 2;1 (G) such that
xt = E (x1 jjFt )
kk
8t 2 [0; 1) ;
then xn !m x as well as x 2 BH , proving the statement.
52
1 < 1.
Remark 6 An important example of M 2;1 (G) is when G is a stopping time -algebra,
that is,
G = fE 2 F : E \ f = tg 2 Ft 8t 2 Ng
where : ! N is a (…nite valued) stopping time. An important bounded A-linear
operator is the map (see [34, p. 391]) f : H ! A such that
8x 2 H:
f (x) = x
Remark 7 We could have de…ned M 2;1 (G) dealing with continuous time. In that case,
we would have needed a …ltration space
; fFt gt2[0;1) ; F; P where F = [t2[0;1) Ft .
With this speci…cation, M 2 would have been the space of cadlag L2 bounded martingales.
Our results would have remained the same.
7
Appendix
Proof of Proposition 1. (i) implies (ii). De…ne l = ' T 1 : C (K) ! R. Since
' and T 1 are both linear and continuous, l 2 C (K). Since ' and T 1 are positive
and such that ' (e) = 1 and T 1 (1K ) = e, l is positive and such that l (1K ) = 1. By
[4, Theorem 14.14], this implies that there exists a (unique) …nite regular probability
measure m on the Borel -algebra of K such that
Z
l (g) =
g (') dm (')
8g 2 C (K) :
K
It follows that
' (a) = l (T (a)) =
Z
T (a) (') dm (') =
K
Z
K
ha; 'i dm (')
8a 2 A:
(29)
By [4, Theorem 12.14] and since m is regular (thus, tight), it follows that suppm exists.
By contradiction, assume that suppm 6= K. Thus, there exists '^ 2 Knsuppm. By
Urysohn’s Lemma (see [4, Theorem 2.46]) and since K is compact and Hausdor¤, there
exists a function 0 g^ 2 C (K) such that g^ (')
^ = 1 and g^ (') = 0 for all ' 2suppm.
Since T is a lattice isomorphism, there exists a
^ 2 A such that T (^
a) = g^ > 0. It follows
that a
^ > 0. Since ' is strictly positive, we can conclude that
Z
Z
0=
h^
a; 'i dm (') =
h^
a; 'i dm (') = ' (^
a) > 0;
suppm
K
a contradiction.
(ii) implies (iii). Since it is immediate to see that K separates the points of A, the
statement trivially follows.
53
(iii) implies (i). It is immediate to see that (1) de…nes a functional ' : A ! R
which is linear, positive, and k kA continuous. Since ' is positive and m a probability
measure, we have that k'kA = ' (e) = 1. We just need to prove that ' is strictly
positive. Consider a 2 A such that a > 0. Since suppm separates the points of A,
it follows that ha; 'i
^ > 0 for some '^ 2 suppm K. This implies that there exists a
nonempty open set V of K such that ha; 'i > 0 for all ' 2 V and m (V ) > 0. We can
conclude that
Z
Z
' (a) =
ha; 'i dm (')
ha; 'i dm (') > 0;
K
V
proving the implication.
1
Proof of the Claim in Proposition 10. Since for each n 2 N, 0
hy; yiH2
1
hy; yiH2 + n1 e = an , it follows that
a2n
hy; yiH
8n 2 N:
2
Since yn = an 1 y for all n 2 N, this implies that (an 1 ) hy; yiH
hyn ; yn iH
e, that is,
8n 2 N:
e
Fix n 2 N. We have that hyn ; yiH = an 1 hy; yiH . First, observe that an
1
implies that an+1
an 1 and hyn ; yiH = an 1 hy; yiH ". Note that
hy; yiH
hy; yiH +
1
1
hy; yiH2 =
n
1
hy; yiH2 +
an+1 . This
1
1
1
e hy; yiH2 = an hy; yiH2 :
n
It follows that
1
hyn ; yiH = an 1 hy; yiH
Also note that
1
1
hy; yiH2
hy; yiH2 +
hy; yiH2 :
1
e = an ;
n
which yields
1
0
an 1 hy; yiH2
e:
Moreover, we can conclude that
1
0
hy; yiH2
1
an 1 hy; yiH = an 1 an hy; yiH2
hy; yiH
1
1 1
an hy; yiH2 :
n
Since k kA is a lattice norm, it follows that
1
hy; yiH2
hyn ; yiH
A
1
1
an 1 hy; yiH2
n
A
1
kekA :
n
Since n was arbitrarily chosen, the statement follows.
Proof of Lemma 3. We …rst show, by steps, that N is a vector-valued norm.
54
Step 1. N (f ) = 0 if and only if f = 0.
Proof of the Step. It is immediate to see that if f = 0, then N (f ) = 0. Viceversa,
note that if N (f ) = 0, then
sup sup
x2H
n2N
jf (x)j
N (x) + n1 e
jf (x)j
0
8x 2 H;
1
n2N N (x) + n e
jf (x)j
0
8n 2 N; 8x 2 H;
=)
N (x) + n1 e
0 =) sup
=) jf (x)j
8x 2 H;
0
=) f (x) = 0
8x 2 H;
proving that f = 0.
Step 2. N (a f ) = jaj N (f ) for all a 2 A and for all f 2 H .
Proof of the Step. Consider a 2 A and f 2 H . It follows that
jaf (x)j
jaj jf (x)j
= sup
sup
1
1
x2H n2N N (x) + n e
x2H
n2N N (x) + n e
jf (x)j
jf (x)j
= jaj sup
= sup jaj sup
sup
1
1
x2H
x2H
n2N N (x) + n e
n2N N (x) + n e
N (a f ) = sup sup
= jaj N (f ) ;
proving the statement.18
Step 3. N (f + g)
N (f ) + N (g) for all f; g 2 H .
Proof of the Step. Consider f; g 2 H . It follows that
j(f + g) (x)j
jf (x) + g (x)j
= sup
sup
1
1
x2H n2N N (x) + n e
x2H
n2N N (x) + n e
jf (x)j + jg (x)j
sup
sup
N (x) + n1 e
x2H
n2N
jf (x)j
jg (x)j
sup
sup
+ sup
1
1
x2H
n2N N (x) + n e
n2N N (x) + n e
jf (x)j
jg (x)j
= sup sup
+ sup sup
= N (f ) + N (g) ;
1
1
x2H n2N N (x) + n e
x2H n2N N (x) + n e
N (f + g) = sup sup
proving the statement.
Steps 1-3 prove that N is a well de…ned vector-valued norm. Finally, observe that
dH (f; g) = 0 () d (0; N (f
() f
g)) = 0 () N (f
g) = 0
g = 0 () f = g:
18
In Step 2, we implicitly used the fact that if a nonempty subset B of A is bounded from above
and c 2 A+ , then sup cB = c sup B (see [11, Footnote 26]).
55
It is immediate to see that dH (f; g) = dH (g; f ) for all f; g 2 H as well as
dH (f + h; g + h) = dH (f; g) for all f; g; h 2 H . Finally, by de…nition of d and
[6, Lemma 1.4] and since N (f + g)
N (f ) + N (g) for all f; g 2 H , we can
conclude that
dH (f; g) = d (0; N (f
' ((N (f
g)) = ' (N (f
h) + N (h
= dH (f; h) + dH (h; g)
g) ^ e) = ' (N ((f
g)) ^ e)
' (N (f
h) + (h
h) ^ e + N (h
g)) ^ e)
g) ^ e)
8f; g; h 2 H ;
proving the statement.
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